Question

please solve this trigonometric identity​

Answer 1

Question :

$\underline{\bf Prove \: the \: following \: identity :}$

• $\bf \dfrac{sin x + cos x}{sin x - cos x} + \dfrac{sin x - cos x}{sin x + cos x} = \dfrac{2}{1-2cos^{2} x}$

Proof :

LHS = $\bf \dfrac{sin x + cos x}{sin x - cos x} + \dfrac{sin x - cos x}{sin x + cos x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{(sin x + cos x)(sin x + cos x) + (sin x - cos x)(sin x - cos x)}{(sin x - cos x)(sin x + cos x)}$

$\sf : \implies \underline{\bf LHS} = \dfrac{(sin x + cos x)^{2} + (sin x - cos x)^{2}}{(sin x - cos x)(sin x + cos x)}$

By using these Algebraic identities :

• (a + b)² = a² + b² + 2ab

• (a - b)² = a² + b² - 2ab

• (a - b)(a + b) = a² - b²

$\sf : \implies \underline{\bf LHS} = \dfrac{(sin^{2} x + cos^{2} x + 2 sin x cos x) + (sin^{2} x + cos^{2} x - 2 sin x cos x)}{sin^{2} x - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{sin^{2} x + cos^{2} x + 2 sin x cos x + sin^{2} x + cos^{2} x - 2 sin x cos x}{sin^{2} x - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{sin^{2} x + cos^{2} x + \cancel{2 sin x cos x} - \cancel{2 sin x cos x} + sin^{2} x + cos^{2} x}{sin^{2} x - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{sin^{2} x + cos^{2} x + sin^{2} x + cos^{2} x}{sin^{2} x - cos^{2} x}$

Now, by using these trigonometric identities :

• sin²θ + cos²θ = 1
• sin²θ = 1 - cos²θ

$\sf : \implies \underline{\bf LHS} = \dfrac{1 + 1}{(1 - cos^{2} x) - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{2}{1 - cos^{2} x - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{2}{1 - 2cos^{2} x}$

RHS = $\bf \dfrac{2}{1-2cos^{2} x}$

$\underline{\boxed{\bf As, \: LHS = RHS,}}$

$\underline{\boxed{\bf Hence, \: proved.}}$

Answer 2

Answer:

Question :

$\underline{\bf Prove \: the \: following \: identity :}$

$\bf \dfrac{sin x + cos x}{sin x - cos x} + \dfrac{sin x - cos x}{sin x + cos x} = \dfrac{2}{1-2cos^{2} x}$

Proof :

LHS =$\bf \dfrac{sin x + cos x}{sin x - cos x} + \dfrac{sin x - cos x}{sin x + cos x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{(sin x + cos x)(sin x + cos x) + (sin x - cos x)(sin x - cos x)}{(sin x - cos x)(sin x + cos x)}$

$\sf : \implies \underline{\bf LHS} = \dfrac{(sin x + cos x)^{2} + (sin x - cos x)^{2}}{(sin x - cos x)(sin x + cos x)}</p><p>By using these Algebraic identities :</p><p></p><p>(a + b)² = a² + b² + 2ab</p><p>(a - b)² = a² + b² - 2ab</p><p>(a - b)(a + b) = a² - b²</p><p>[tex]\sf : \implies \underline{\bf LHS} = \dfrac{(sin^{2} x + cos^{2} x + 2 sin x cos x) + (sin^{2} x + cos^{2} x - 2 sin x cos x)}{sin^{2} x - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{sin^{2} x + cos^{2} x + 2 sin x cos x + sin^{2} x + cos^{2} x - 2 sin x cos x}{sin^{2} x - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{sin^{2} x + cos^{2} x + \cancel{2 sin x cos x} - \cancel{2 sin x cos x} + sin^{2} x + cos^{2} x}{sin^{2} x - cos^{2} x}$

$\sf : \implies \underline{\bf LHS} = \dfrac{sin^{2} x + cos^{2} x + sin^{2} x + cos^{2} x}{sin^{2} x - cos^{2} x}$

Now, by using these trigonometric identities :

sin²θ + cos²θ = 1

sin²θ = 1 - cos²θ

\sf : \implies \underline{\bf LHS} = \dfrac{1 + 1}{(1 - cos^{2} x) - cos^{2} x}:

$\sf : \implies \underline{\bf LHS} = \dfrac{2}{1 - cos^{2} x - cos^{2} x}:$

$\sf : \implies \underline{\bf LHS} = \dfrac{2}{1 - 2cos^{2} xx$

RHS =$\bf \dfrac{2}{1-2cos^{2} x}$

$\underline{\boxed{\bf As, \: LHS = RHS,}}$

$\underline{\boxed{\bf Hence, \: proved.}}$