Problem and Solution - Proving Trigonometric Identities

Here we will prove the problems on trigonometric identities. As you know that the identity consists of two sides in equation, named Left Hand Side (abbreviated as LHS) and Right Hand Side (abbreviated as RHS). To prove the identity, sometimes we need to apply more fundamental identities, e.g. $\sin^2 x + \cos^2 x = 1$ and use logical steps in order to lead one side form into the other side of the equation. Usually we start the proving process from the more complex one first. Leading to this, of course, we never say it easy to do. But once you keep trying to exercise, you will learn it “automagically”, we might say.

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Problem Number 1
Prove the following trigonometric identity.
$\dfrac{2 -\sec^2 A} {\sec^2 A} = 1 -2 \sin^2 A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \cos x & = \dfrac{1}{\sec x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} \dfrac{2 -\sec^2 A} {\sec^2 A} & = \dfrac{2}{\sec^2 A}- \dfrac{\sec^2 A} {\sec^2 A} \\ & = 2(\cos^2 A) -1 \\ & = 2(1 -\sin^2 A) -1 \\ & = 1 -2 \sin^2 A \end{aligned}$
Thus, it is proved that $\dfrac{2 -\sec^2 A} {\sec^2 A} = 1 -2 \sin^2 A$

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Problem Number 2
Prove the following trigonometric identity.
$$(\sin A + \cos A)^2 -(\sin A -\cos A)^2 = 4 \sin A \cos A$$

Proof

The proof is started from the left-hand side:
$\begin{aligned} & (\sin A + \cos A)^2 -(\sin A -\cos A)^2 \\ =& (\cancel{\sin^2 A} + 2 \sin A \cos A + \bcancel{\cos^2 A} ) \\ & -(\cancel{\sin^2 A} -2 \sin A \cos A + \bcancel{\cos^2 A}) \\ = & 2 \sin A \cos A -(-2 \sin A \cos A) \\ = & 4 \sin A \cos A \end{aligned}$
Thus, it is proved that $$(\sin A + \cos A)^2 -(\sin A -\cos A)^2 = 4 \sin A \cos A$$

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Problem Number 3
Prove the following trigonometric identity.
$\dfrac{\sec^2 \theta -1}{\sec^2 \theta} = \sin^2 \theta$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x &= 1 && (\text{Pythagorean Identity}) \\ \dfrac{1}{\sec x} & = \cos x && (\text{Reciprocal Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} \dfrac{\sec^2 \theta -1}{\sec^2 \theta} & = \dfrac{\sec^2 \theta} {\sec^2 \theta} -\dfrac{1}{\sec^2 \theta} \\ & = 1 -\cos^2 \theta \\ & = \sin^2 \theta \end{aligned}$
Thus, it is proved that $\dfrac{\sec^2 \theta -1}{\sec^2 \theta} = \sin^2 \theta$

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Problem Number 4
Prove the following trigonometric identity.
$\dfrac{\sin A} {1 -\cos A} = \dfrac{1 + \cos A} {\sin A}$

Proof

We will apply the following more fundamental trigonometric identities: adalah Identitas Pythagoras, yaitu
$\boxed{\sin^2 x + \cos x = 1}$
The proof is started from the left-hand side.:
$\begin{aligned} \dfrac{\sin A} {1 -\cos A} & = \dfrac{\sin A} {1 -\cos A} \times \dfrac{1 + \cos A} {1 + \cos A} \\ & = \dfrac{\sin A(1 + \cos A)} {1 -\cos^2 A} \\ & = \dfrac{\cancel{\sin A}(1 + \cos A)} {\sin^{\cancel{2}} A} \\ & = \dfrac{1 + \cos A} {\sin A} \end{aligned}$
Thus, it is proved that $\dfrac{\sin A} {1 -\cos A} = \dfrac{1 + \cos A} {\sin A}$

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Problem Number 5
Prove the following trigonometric identity.
$\sin A + \cos A \cot A = \csc A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$$\begin{aligned} \sin A + \cos A \cot A & = \sin A + \cos A \left(\dfrac{\cos A} {\sin A}\right) \\ & = \sin A + \dfrac{\cos^2 A} {\sin A} \\ & = \dfrac{\sin^2 A + \cos^2 A} {\sin A} \\ & = \dfrac{1}{\sin A} = \csc A \end{aligned}$$Thus, it is proved that $\sin A + \cos A \cot A = \csc A$

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Problem Number 6
Prove the following trigonometric identity.
$\csc A + \cot A = \dfrac{\sin A} {1- \cos A}$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the right-hand side.
$\begin{aligned} \dfrac{\sin A} {1- \cos A} & = \dfrac{\sin A} {1-\cos A} \times \dfrac{1 + \cos A} {1 + \cos A} \\ & = \dfrac{\sin A(1 + \cos A)} {1 -\cos^2 A} \\ & = \dfrac{\cancel{\sin A}(1 + \cos A)} {\sin^{\cancel{2}} A} \\ & = \dfrac{1 + \cos A} {\sin A} \\ & = \csc A + \cot A \end{aligned}$
Thus, it is proved that $\csc A + \cot A = \dfrac{\sin A} {1- \cos A}$

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Problem Number 7
Prove the following trigonometric identity.
$\sin A \csc A -\sin^2 A = \cos^2 A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cos^2 x + \sin^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.:
$\begin{aligned} \sin A \csc A -\sin^2 A & = \sin A \cdot \dfrac{1}{\sin A} -\sin^2 A \\ & = 1 -\sin^2 A = \cos^2 A \end{aligned}$
Thus, it is proved that $\sin A \csc A -\sin^2 A = \cos^2 A$

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Problem Number 8
Prove the following trigonometric identity.
$(\csc A + \cot A)(1 -\cos A) = \sin A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cot x & = \dfrac{\cos x}{\sin x} && (\text{Quotient Identity}) \\ \cos^2 x + \sin^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} & (\csc A + \cot A)(1 -\cos A) \\ & = \csc A -\csc A \cos A + \cot A -\cot A \cos A \\ & = \dfrac{1}{\sin A} -\dfrac{\cos A}{\sin A} + \cot A- \dfrac{\cos A}{\sin A} \cdot \cos A \\ & = \dfrac{1}{\sin A} – \cot A + \cot A- \dfrac{\cos^2 A}{\sin A} \\ & = \dfrac{1 -\cos^2 A}{\sin A} \\ & = \dfrac{\sin^2 A}{\sin A} = \sin A \end{aligned}$
Thus, it is proved that $(\csc A + \cot A)(1 -\cos A) = \sin A$

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Problem Number 9
Prove the following trigonometric identity.
$\tan^2 A -\sin^2 A = \tan^2 A \sin^2 A$

Proof

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Problem Number 10
Prove the following trigonometric identity.
$\tan A \cos^4 A + \cot A \sin^4 A = \sin A \cos A$

Proof

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Problem Number 11
Prove the following trigonometric identity.
$\dfrac{\sin^2 A} {\cos^2 A} -\dfrac{\cos^2 A} {\sin^2 A} = \sec^2 A -\csc^2 A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} & \dfrac{\sin^2 A} {\cos^2 A} -\dfrac{\cos^2 A} {\sin^2 A} \\ & = \dfrac{\cos^4 A -\sin^4 A} {\cos^2 A \sin^2 A} \\ & = \dfrac{(\cos^2 A -\sin^2 A)(\cos^2 A + \sin^2 A)} {\cos^2 A \sin^2 A} \\ & = \dfrac{(\cos^2 A -\sin^2 A)(1)} {\cos^2 A~\sin^2 A} \\ & = \dfrac{\cancel{\cos^2 A}} {\cancel{\cos^2 A} ~\sin^2 A} -\dfrac{\bcancel{\sin^2 A}} {\cos^2 A~\bcancel{\sin^2 A}} \\ & = \dfrac{1}{\cos^2 A} -\dfrac{1}{\sin^2 A} \\ & = \sec^2 A -\csc^2 A \end{aligned}$
Thus, it is proved that $\dfrac{\sin^2 A} {\cos^2 A} -\dfrac{\cos^2 A} {\sin^2 A} = \sec^2 A -\csc^2 A$

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Problem Number 12
Prove the following trigonometric identity.
$\dfrac{\sin^2 A -\sin^2 B} {\cos^2 A \cos^2 B} = \tan^2 A -\tan^2 B$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \sec^2 x & = 1 + \tan^2 x && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$$\begin{aligned} & \dfrac{\sin^2 A -\sin^2 B} {\cos^2 A \cos^2 B} \\ & = \dfrac{\sin^2 A}{\cos^2 A \cos^2 B} -\dfrac{\sin^2 B} {\cos^2 A \cos^2 B}\\ & = \dfrac{\sin^2 A} {\cos^2 A} \cdot \dfrac{1}{\cos^2 B} -\dfrac{\sin^2 B} {\cos^2 B} \cdot \dfrac{1}{\cos^2 A} \\ & = \tan^2 A(\sec^2 B) -\tan^2 B(\sec^2 A) \\ & = \tan^2 A(1 + \tan^2 B) -\tan^2 B(1 + \tan^2 A) \\ & = \tan^2 A + \cancel{\tan^2 A \tan^2 B} -\tan^2 B -\cancel{\tan^2 A \tan^2 B} \\ & = \tan^2 A- \tan^2 B \end{aligned}$$Thus, it is proved that $\dfrac{\sin^2 A -\sin^2 B} {\cos^2 A \cos^2 B} = \tan^2 A- \tan^2 B$

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Problem Number 13
Prove the following trigonometric identity.
$\cos^2 A + \cot^2 A \cos^2 A = \cot^2 A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} 1 + \cot^2 x & = \csc^2 x && (\text{Pythagorean Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} & \cos^2 A + \cot^2 A \cos^2 A \\ & = \cos^2 A(1 + \cot^2 A) \\ & = \cos^2 A(\csc^2 A) \\ & = \cos^2 A \cdot \dfrac{1}{\sin^2 A} \\ & = \cot^2 A \end{aligned}$
Thus, it is proved that $\cos^2 A + \cot^2 A \cos^2 A = \cot^2 A$

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Problem Number 14
Prove the following trigonometric identity
$\dfrac{1 + \sec A} {\tan A + \sin A} = \csc A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} \dfrac{1 + \sec A} {\tan A + \sin A} & = \dfrac{1 + \frac{1}{\cos A}} {\frac{\sin A} {\cos A} + \sin A} \\ & = \dfrac{\cos A + 1}{\sin A + \sin A \cos A} \\ & = \dfrac{\cancel{\cos A + 1}} {\sin A(\cancel{1 + \cos A}) } \\ & = \dfrac{1}{\sin A} = \csc A \end{aligned}$
Thus, it is proved that $\dfrac{1 + \sec A} {\tan A + \sin A} = \csc A$

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Problem Number 15
Prove the following trigonometric identity.
$\dfrac{1 + \sin x} {1 -\sin x} = (\sec x + \tan x)^2$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} \dfrac{1 + \sin x} {1 -\sin x} & = \dfrac{1 + \sin x} {1 -\sin x} \times \dfrac{1 + \sin x} {1 + \sin x} \\ & = \dfrac{1 + 2 \sin x + \sin^2 x} {1- \sin^2 x} \\ & = \dfrac{1 + 2 \sin x + \sin^2 x} {\cos ^2 x} \\ & = \dfrac{1}{\cos^2 x} + \dfrac{2 \sin x} {\cos^2 x} + \dfrac{\sin^2 x} {\cos^2 x} \\ & = \sec^2 x + 2 \sec x \tan x + \tan^2 x \\ & = (\sec x + \tan x)^2 \end{aligned}$
Thus, it is proved that $\dfrac{1 + \sin x} {1 -\sin x} = (\sec x + \tan x)^2$

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Problem Number 16
Prove the following trigonometric identity.
$\tan A = \dfrac12 \sin 2A(1+\tan^2 A)$

Proof

We will apply the following more fundamental trigonometric identities:
$\boxed{\begin{aligned} \sin 2x & = 2 \sin x \cos x \\ \tan^2 x + 1 & = \sec^2 x \end{aligned}}$
The proof is started from the right-hand side.
$\begin{aligned} & \dfrac12 \sin 2A(1+\tan^2 A) \\ & = \dfrac12(2 \sin A \cos A) (1+\tan^2 A) \\ & = (\sin A \cos A) (1+\tan^2 A) \\ & = (\sin A \cos A) (\sec^2 A) \\ & = \dfrac{\sin A \cancel{\cos A}} {\cancelto{\cos A} {\cos^2 A}} \\ & = \dfrac{\sin A} {\cos A} = \tan A \end{aligned}$
Thus, it is proved that $\tan A = \dfrac12 \sin 2A(1+\tan^2 A)$

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Problem Number 17
Prove the following trigonometric identity.
$\tan^2 A = 1 -\cos 2A(1 + \tan^2 A)$

Proof

We will apply the following more fundamental trigonometric identities:
$\boxed{\begin{aligned} 1+\tan^2 x & = \sec^2 x \\ \cos 2x & = 1 -2 \sin^2 x \\ \sec x & = \dfrac{1}{\cos x} \end{aligned}}$
The proof is started from the right-hand side.
$\begin{aligned} & 1 – \cos 2A(1 + \tan^2 A) \\ & = 1 -\cos 2A(\sec^2 A) \\ & = 1- \dfrac{1 -2 \sin^2 A}{\cos^2 A} \\ & = 1 -\dfrac{1}{\cos^2 A} + \dfrac{2 \sin^2 A} {\cos^2 A} \\ & = 1 -\sec^2 A + 2 \tan^2 A \\ & = 1 -(1+\tan^2 A) + 2 \tan^2 A \\ & = \tan^2 A \end{aligned}$
Thus, it is proved that $\tan^2 A = 1 -\cos 2A(1 + \tan^2 A)$

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Problem Number 18
Prove the following trigonometric identity.
$$\tan^6 A = \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \sec^2 A -1$$

Proof

We will apply the following more fundamental trigonometric identity:
$\boxed{\tan^2 x + 1 = \sec^2 x}$
The proof is started from the right-hand side.
$$\begin{aligned} & \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \sec^2 A -1 \\ & = \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \tan^2 A \\ & = \tan^2 A(\tan^2 A \sec^2 A -\sec^2 A + 1) \\ & = \tan^2 A((\sec^2 A)(\tan^2 A- 1)+1) \\ & = \tan^2 A((\tan^2 A + 1)(\tan^2 A -1)+1) \\ & = \tan^2 A(\tan^4 A-1 + 1) \\ & = \tan^6 A \end{aligned}$$Thus, it is proved that $$\tan^6 A = \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \sec^2 A -1$$

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Problem Number 19
Prove the following trigonometric identity.
$\tan (A -B) = \dfrac{\sin 2A -\sin 2B} {\cos 2A + \cos 2B}$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \sin x -\sin y & = 2 \cos \dfrac12(x + y) \sin \dfrac12(x-y) \\ \cos x + \cos y & = 2 \cos \dfrac12(x + y) \cos \dfrac12(x-y) \\ \tan x & = \dfrac{\sin x} {\cos x} \end{aligned}}$$The proof is started from the right-hand side.
$$\begin{aligned} \dfrac{\sin 2A -\sin 2B} {\cos 2A + \cos 2B} & = \dfrac{2 \cos \dfrac12(2A + 2B) \sin \dfrac12(2A-2B)} {2 \cos \dfrac12(2A+2B) \cos \dfrac12(2A-2B)} \\ & = \dfrac{\cancel{2 \cos (A+B)} \sin (A-B)} {\cancel{2 \cos (A+B)} \cos (A-B)} \\ & = \dfrac{\sin (A-B)} {\cos (A-B)} = \tan (A-B) \end{aligned}$$Thus, it is proved that $\tan (A -B) = \dfrac{\sin 2A -\sin 2B} {\cos 2A + \cos 2B}$

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Problem Number 20
Prove the following trigonometric identity.
$\cot 2A = \dfrac{\sin A -\sin 3A} {\cos 3A -\cos A}$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \sin x -\sin y & = 2 \cos \dfrac12(x + y) \sin \dfrac12(x-y) \\ \cos x -\cos y & = -2 \sin \dfrac12(x + y) \sin \dfrac12(x-y) \\ \sin (-x) & = -\sin x \\ \cot x & = \dfrac{\cos x} {\sin x} \end{aligned}}$$The proof is started from the right-hand side.:
$$\begin{aligned} \dfrac{\sin A -\sin 3A} {\cos 3A -\cos A} & = \dfrac{2 \cos \dfrac12(A + 3A) \sin \dfrac12(A-3A)} {-2 \sin \dfrac12(3A + A) \sin \dfrac12(3A- A)} \\ & = \dfrac{2 \cos 2A \sin (-A)} {-2 \sin 2A \sin A} \\ & = \dfrac{\cancel{-2 \sin A} \cos 2A} {\cancel{-2 \sin A} \sin 2A} \\ & = \dfrac{\cos 2A} {\sin 2A} = \cot 2A \end{aligned}$$Thus, it is proved that $\cot 2A = \dfrac{\sin A -\sin 3A} {\cos 3A -\cos A}$

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Problem Number 21
Prove the following trigonometric identity.
$\dfrac{1 -\cos 2A + \sin A} {\sin 2A + \cos A} =\tan A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \cos 2x & = \cos^2 x – \sin^2 x && (\text{Double Angle Identity}) \\ \sin 2x &= 2 \sin x \cos x && (\text{Double Angle Identity}) \\ \sin^2 x + \cos^2 x &= 1 && (\text{Pythagorean Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$$\begin{aligned} \dfrac{1 -\cos 2A + \sin A} {\sin 2A + \cos A} & = \dfrac{1- (\cos^2 A -\sin^2 A) + \sin A} {(2 \sin A \cos A) + \cos A} \\ & = \dfrac{(1- \cos^2 A) + \sin^2 A + \sin A} {\cos A(2 \sin A + 1)} \\ & = \dfrac{\sin^2 A + \sin^2 A + \sin A} {\cos A(2 \sin A + 1)} \\ & = \dfrac{\sin A\cancel{(2 \sin A + 1)}} {\cos A\cancel{(2 \sin A + 1)}} \\ & = \dfrac{\sin A} {\cos A} = \tan A \end{aligned}$$Thus, it is proved that $\dfrac{1 -\cos 2A + \sin A} {\sin 2A + \cos A} =\tan A$

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Problem Number 22
Prove the following trigonometric identity.
$\csc 2A + \cot 2A = \cot A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \cos 2x &= \cos^2 x -\sin^2 x && (\text{Double Angle Identity}) \\ \sin 2x & = 2 \sin x \cos x && (\text{Double Angle Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} \csc 2A + \cot 2A & = \dfrac{1}{\sin 2A} + \dfrac{\cos 2A} {\sin 2A} \\ & = \dfrac{1 + (\cos^2 A -\sin^2 A)} {2 \sin A \cos A} \\ & = \dfrac{(1- \sin^2 A) + \cos^2 A} {2 \sin A \cos A} \\ & = \dfrac{2 \cos^2 A} {2 \sin A \cos A} \\ & = \dfrac{\cos A} {\sin A} = \cot A \end{aligned}$
Thus, it is proved that $\csc 2A + \cot 2A = \cot A$.

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Problem Number 23
Prove the following trigonometric identity.
$\dfrac{2 \sin (A-B)} {\cos (A-B)-\cos (A+B)} = \cot B -\cot A$

Proof

Recall the addition and subtraction identities:
$\boxed{\begin{aligned} \sin (x \pm y) & = \sin x \cos y \pm \cos x \sin y \\ \cos (x \pm y) & = \cos x \cos y \mp \sin x \sin y \end{aligned}}$
Additionally, we will apply the following more fundamental trigonometric identities:
$ \cot x = \dfrac{\cos x} {\sin x} $
The proof is started from the left-hand side.
$$\begin{aligned} & \dfrac{2 \sin (A-B)} {\cos (A-B)-\cos (A+B)} \\ & = \dfrac{2 (\sin A \cos B -\sin B \cos A)} {(\cancel{\cos A \cos B} + \sin A \sin B)- (\cancel{\cos A \cos B} -\sin A \sin B)} \\ & = \dfrac{\cancel{2}(\sin A \cos B -\sin B \cos A)} {\cancel{2} \sin A \sin B} \\ & = \dfrac{\sin A \cos B -\sin B \cos A} {\sin A \sin B} \\ & = \dfrac{\cancel{\sin A} \cos B} {\cancel{\sin A} \sin B}- \dfrac{\cancel{\sin B} \cos A} {\sin A \cancel{\sin B}} \\ & = \dfrac{\cos B} {\sin B} -\dfrac{\cos A} {\sin A} \\ & = \cot B -\cot A \end{aligned}$$Thus, it is proved that $\dfrac{2 \sin (A-B)} {\cos (A-B)-\cos (A+B)} = \cot B -\cot A$

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Problem Number 24
Prove the following trigonometric identity.
$\tan A -\cot A = \dfrac{1 -2 \sin^2 A} {\sin A \cos A}$

Proof

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Problem Number 25
Prove the following trigonometric identity.
$1 -\tan^2 A= \cos 2A \sec^2 A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \cos 2x & = \cos^2 x -\sin^2 x && (\text{Double Angle Identity}) \\ \sec x & =\dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$The proof is started from the right-hand side.
$\begin{aligned} \cos 2A \sec^2 A & = (\cos^2 A -\sin^2 A) \cdot \dfrac{1}{\cos^2 A} \\ & = 1 -\dfrac{\sin^2 A} {\cos^2 A} \\ & = 1 -\tan^2 A \end{aligned}$
Thus, it is proved that $1 -\tan^2 A= \cos 2A \sec^2 A.$

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Problem Number 26
Prove the following trigonometric identity.
$\dfrac{\sin \left(A + \frac{\pi} {4}\right)} {\cos \left(A + \frac{\pi} {4}\right)} + \dfrac{\cos \left(A + \frac{\pi} {4}\right)} {\sin \left(A + \frac{\pi} {4}\right)} = 2 \sec 2A$

Proof

Recall the following supplement angle identity:
$\sin (2x + 90^{\circ}) = \cos 2x$
Furthermore, we will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \sin 2x & = 2 \sin x \cos x && (\text{Double Angle Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} & \dfrac{\sin \left(A + \frac{\pi} {4}\right)} {\cos \left(A + \frac{\pi} {4}\right)} + \dfrac{\cos \left(A + \frac{\pi} {4}\right)} {\sin \left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{\sin^2 \left(A + \frac{\pi} {4}\right) + \cos^2 \left(A + \frac{\pi} {4}\right)}{\sin \left(A + \frac{\pi} {4}\right) \cdot \cos \left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{1}{\sin \left(A + \frac{\pi} {4}\right) \cdot \cos \left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{2}{\sin 2\left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{2}{\sin \left(2A + \frac{\pi} {2}\right)} \\ & = \dfrac{2}{\cos 2A} = 2 \sec 2A \end{aligned}$
Thus, it is proved that $\dfrac{\sin \left(A + \frac{\pi} {4}\right)} {\cos \left(A + \frac{\pi} {4}\right)} + \dfrac{\cos \left(A + \frac{\pi} {4}\right)} {\sin \left(A + \frac{\pi} {4}\right)} = 2 \sec 2A$

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Problem Number 27
Prove the following trigonometric identity.
$\dfrac{\sin 4A} {1 + \cos 4A} = \tan 2A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \sin 2x & = 2 \sin x \cos x && (\text{Double Angle Identity}) \\ \cos 2x & = \cos^2 x -\sin^2 x && (\text{Double Angle Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} \dfrac{\sin 4A} {1 + \cos 4A} & = \dfrac{2 \sin 2A \cos 2A} {1 + (\cos^2 2A -\sin^2 2A)} \\ & = \dfrac{2 \sin 2A \cos 2A} {(1 -\sin^2 2A) + \cos^2 2A} \\ & = \dfrac{2 \sin 2A \cos 2A} {2 \cos^2 2A} \\ & = \dfrac{\sin 2A} {\cos 2A} = \tan 2A \end{aligned}$
Thus, it is proved that $\dfrac{\sin 4A} {1 + \cos 4A} = \tan 2A$

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Problem Number 28
Prove the following trigonometric identity.
$\begin{aligned} (\cos 2a + \cos 4a & + \cos 6a) \sin a \\ & = \cos 4a \sin 3a \end{aligned}$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \cos A + \cos B & = 2 \cos \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A-B}{2}\right) \\ \cos A \sin B & = \dfrac12\left(\sin (A + B) + \sin (A-B)\right) \end{aligned}}$$The proof is started from the left-hand side.:
$$\begin{aligned} & (\cos 2a + \cos 4a + \cos 6a) \sin a \\ & = \left(\cos 4a + 2 \cos \left(\dfrac{2a+6a}{2}\right) \cos \left(\dfrac{6a-2a}{2}\right)\right) \sin a \\ & = (\cos 4a + 2 \cos 4a \cos 2a) \sin a \\ & = \cos 4a(1 + 2 \cos 2a) \sin a \\ & = \cos 4a(\sin a + 2 \cos 2a \sin a) \\ & = \cos 4a(\sin a + \cancel{2} \cdot \dfrac{1}{\cancel{2}}(\sin (a+2a) + \sin (a-2a))) \\ & = \cos 4a(\cancel{\sin a} \sin 3a~ \cancel{\sin (-a)}) \\ & = \cos 4a \sin 3a \end{aligned}$$Thus, it is proved that
$\begin{aligned} (\cos 2a + \cos 4a & + \cos 6a) \sin a \\ & = \cos 4a \sin 3a \end{aligned}$

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Problem Number 29
Prove the following trigonometric identity.
$2 \sin A \cos (A + 30^{\circ}) = \sin(2A + 30^{\circ}) -\dfrac12$

Proof

We will apply the following more fundamental trigonometric identities:
$\boxed{\begin{aligned} \cos (A + B) & = \cos A \cos B – \sin A \sin B \\ \sin 2A & = 2 \sin A \cos A \\ 1 -\cos 2A & = 2 \sin^2 A \\ \sin(A + B) & = \sin A \cos B + \sin B \cos A \end{aligned}}$
The proof is started from the left-hand side.
$\begin{aligned} & 2 \sin A \cos (A + 30^{\circ}) \\ & = 2 \sin A(\cos A \cos 30^{\circ} -\sin A \sin 30^{\circ}) \\ & = 2 \sin A\left(\dfrac12\sqrt3 \cos A- \dfrac12 \sin A\right) \\ & = \dfrac12\sqrt3(2 \sin A \cos A) -\dfrac12(\sin^2 A) \\ & = \dfrac12\sqrt3(\sin 2A) -\dfrac12(1 -\cos 2A) \\ & = \dfrac12\sqrt3(\sin 2A)- \dfrac12 + \dfrac12 \cos 2A \\ & = (\sin 2A \cos 30^{\circ} + \cos 2A \sin 30^{\circ}) -\dfrac12 \\ & = \sin (2A + 30^{\circ}) -\dfrac12 \end{aligned}$
Thus, it is proved that
$2 \sin A \cos (A + 30^{\circ}) = \sin(2A + 30^{\circ}) -\dfrac12$

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Problem Number 30
Prove the following trigonometric identity.
$2 \cos \left(\dfrac{\pi}{4} + A\right) \cos \left(\dfrac{3\pi}{4} -A \right) = -1 +\sin 2A$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} 2 \cos A \cos B & = \cos (A+B) + \cos (A-B) \\ \sin x & = \cos (90^{\circ} -x) = \cos (x -90^{\circ}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} & 2 \cos \left(\dfrac{\pi}{4} + A\right) \cos \left(\dfrac{3\pi}{4}- A \right) \\ & = \cos \left(\left(\dfrac{\pi}{4}+A\right) + \left(\dfrac{3\pi}{4}- A\right)\right) \\ & + \cos \left(\left(\dfrac{\pi}{4} + A\right) -\left(\dfrac{3\pi}{4} – A\right)\right) \\ & = \cos \pi + \cos \left(-\dfrac{\pi}{2} + 2A\right) \\ & = -1 + \sin 2A \end{aligned}$
Thus, it is proved that $$2 \cos \left(\dfrac{\pi}{4} + A\right) \cos \left(\dfrac{3\pi}{4} -A \right) = -1 +\sin 2A$$

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Problem Number 31
Prove the following trigonometric identity.
$\begin{aligned} \sin^4 x -\cos^4 x & = 1 -2(\sin x \cos x)^2\\ & -2 \cos^4 x \end{aligned}$

Proof

We will apply the Pythagorean identity:
$\boxed{\sin^2 x + \cos^2 x = 1}$
The proof is started from the right-hand side.
$$\begin{aligned} & 1 -2(\sin x \cos x)^2 – 2 \cos^4 x \\ & = 1 -2 \sin^2 x \cos^2 x -2 \cos^4 x \\ & = (\sin^2 x + \cos^2 x) -2(1- \cos^2 x)(\cos^2 x) -2 \cos^4 x \\ & = (\sin^2 x + \cos^2 x) -2 \cos^2 x + \cancel{2 \cos^4 x -2 \cos^4 x} \\ & = \sin^2 x -\cos^2 x \\ & = (\sin^2 x -\cos^2 x)(1) \\ & = (\sin^2 x -\cos^2 x) (\sin^2 x + \cos^2 x) \\ & = \sin^4 x -\cos^4 x \end{aligned}$$Thus, it is proved that $\begin{aligned} \sin^4 x -\cos^4 x & = 1 -2(\sin x \cos x)^2\\ & -2 \cos^4 x \end{aligned}$

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Problem Number 32
Prove the following trigonometric identity.
$\cos^6 x + 3 \sin^2 x \cos^2 x + \sin^6 x = 1$

Proof

We will apply the following more fundamental trigonometric identities and the expansion of the third power of a binomial.
$$\boxed{\begin{aligned} (a+b)^3 & = a^3 + 3a^2b + 3ab^2 + b^3 \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the right-hand side.
$$\begin{aligned} 1 & = 1^3 \\ & = (\sin^2 x + \cos^2 x)^3 \\ & = \sin^6 x + \cos^6 x + 3 \sin^4 x \cos^2 x + 3 \sin^2 x \cos^4 x \\ & = \sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x(\sin^2 x + \cos^2 x) \\ & = \sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x(1) \\ & = \cos^6 x + 3 \sin^2 x \cos^2 x + \sin^6 x \end{aligned}$$Thus, it is proved that $\cos^6 x + 3 \sin^2 x \cos^2 x + \sin^6 x = 1$

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Problem Number 33
Prove the following trigonometric identity.
$\dfrac{\tan x-\sin x}{\sin^3 x} = \dfrac{1}{\cos x(1 + \cos x)}$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x}{\cos x} && (\text{Quotient Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$$\begin{aligned} \dfrac{\tan x-\sin x}{\sin^3 x} & = \dfrac{\dfrac{\sin x}{\cos x}- \sin x}{\sin^3 x} \times \color{red}{\dfrac{\cos x}{\cos x}} \\ & = \dfrac{\sin x -\sin x \cos x}{\sin^3 x \cos x} \\ & = \dfrac{\cancel{\sin x}(1 -\cos x)}{\cancelto{\sin^2 x}{\sin^3 x} \cos x} \\ & = \dfrac{1-\cos x}{\sin^2 x \cos x} \\ & = \dfrac{1- \cos x}{(1 -\cos^2 x) \cos x} \\ & = \dfrac{\cancel{1- \cos x}}{\cancel{(1 -\cos x)}(1 + \cos x)\cos x} \\ & = \dfrac{1}{\cos x(1 + \cos x)} \end{aligned}$$Thus, it is proved that $\dfrac{\tan x- \sin x}{\sin^3 x} = \dfrac{1}{\cos x(1 + \cos x)}$

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Problem Number 34
Prove the following trigonometric identity.
$\begin{aligned} & \dfrac{\cot x-\tan x}{(\cos x-\sin x)(\cot x+\tan x)} \\ & = \sin x + \cos x \end{aligned}$

Proof

We will apply the following more fundamental trigonometric identities:
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x}{\cos x} && (\text{Quotient Identity}) \\ \cot x & = \dfrac{\cos x}{\sin x} && (\text{Quotient Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$The proof is started from the left-hand side.
$\begin{aligned} & \dfrac{\cot x-\tan x}{(\cos x-\sin x)(\cot x+\tan x)} \\ & = \dfrac{\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}}{(\cos x-\sin x)\left(\dfrac{\cos x}{\sin x}+ \dfrac{\sin x}{\cos x}\right)} \\ & = \dfrac{\dfrac{cos^2 x- \sin^2 x}{\cancel{\sin x \cos x}}}{(\cos x-\sin x)\left(\dfrac{\cos^2 x + \sin^2 x}{\cancel{\sin x \cos x}}\right)} \\ & = \dfrac{\cos^2 x- \sin^2 x}{(\cos x-\sin x)(\sin^2 x + \cos^2 x)} \\ & = \dfrac{(\cos x + \sin x)\cancel{(\cos x- \sin x)}}{\cancel{(\cos x-\sin x)}(1)} \\ & = \sin x + \cos x \end{aligned}$
Thus, it is proved that $\begin{aligned} & \dfrac{\cot x-\tan x}{(\cos x-\sin x)(\cot x+\tan x)} \\ & = \sin x + \cos x \end{aligned}$

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Problem Number 35
Prove the following trigonometric identity.
$\dfrac{(\tan x-1)(\tan x + \cot x)}{\tan x- \cot x} = \dfrac{\tan^2 x+1}{\tan x +1}$

Proof

We will apply the following more fundamental trigonometric identity:
$\boxed{\tan x \cot x = 1~~~~~(\text{Reciprocal Identity})}$
The proof is started from the right-hand side.:
$$\begin{aligned} \dfrac{\tan^2 x +1}{\tan x +1} & = \dfrac{\tan^2 x + \tan x \cot x}{\tan x + \tan x \cot x} \\ & = \dfrac{\cancel{\tan x}(\tan x + \cot x)}{\cancel{\tan x}(1 + \cot x)} \\ & = \dfrac{\tan x + \cot x}{1+\cot x} \times \color{red}{\dfrac{\tan x-\cot x}{\tan x-\cot x}} \\ & = \dfrac{(\tan x + \cot x)(\tan x -\cot x)}{(\tan x-\cot x)(1+\cot x)} \times \color{blue}{\dfrac{\tan x- 1}{\tan x-1}} \\ & = \dfrac{(\tan x + \cot x)(\tan x-\cot x)(\tan x-1)}{(\tan x-\cot x)(\tan x \cancel{-1 + \tan x \cos x} -\cot x)} \\ & = \dfrac{(\tan x + \cot x)\cancel{(\tan x-\cot x)}(\tan x-1)}{(\tan x-\cot x)\cancel{(\tan x – \cot x)}} \\ & = \dfrac{(\tan x -1)(\tan x + \cot x)}{\tan x-\cot x} \end{aligned}$$Thus, it is proved that $\dfrac{(\tan x-1)(\tan x + \cot x)}{\tan x- \cot x} = \dfrac{\tan^2 x+1}{\tan x +1}$

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Problem Number 36
Show that $\sin^6 A + \cos^6 A + \dfrac34 \sin^2 2A = 1$.

Proof

Recall the expansion of the third power of a binomial.
$\begin{aligned} (a+b)^3 & = a^3+b^3+3a^2b+3ab^2 \\ \Rightarrow (a^2 + b^2)^3 & = a^6 + b^6+3a^4b^2+3a^2b^4 \\ & =a^6+b^6+3a^2b^2(a^2+b^2) \end{aligned}$
Additionally, we will apply the following identities.
$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 \\ \sin 2x & = 2 \sin x \cos x \end{aligned}}$
Starting the proof from the left-hand side, we will have
$$\begin{aligned} &\sin^6 A + \cos^6 A + \dfrac34 \sin^2 2A \\ & = (\sin^2 A + \cos^2 A)^3-3 \sin^2 A \cos^2 A(\sin^2 A + \cos^2 A)+\dfrac34((2 \sin A \cos A)^2) \\ & = 1^3-3 \sin^2 A \cos^2 A(1) + \dfrac{3}{\cancel{4}}(\cancel{4} \sin^2 A \cos^2 A) \\ & = 1-\cancel{3 \sin^2 A \cos^2 A}+\cancel{3 \sin^2 A \cos^2 A} \\ & = 1 \end{aligned}$$Thus, it is proved that $\sin^6 A + \cos^6 A + \dfrac34 \sin^2 2A = 1$.

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