Question

Heres my question

Trigonometry

Class 10...

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Answer 1

To Prove

(cosx + 1)/sin³x = cosecx/(1 - cosx)

Proof

Solving R.H.S we get,

→ cosecx/(1 - cosx)

Multiplying it with (1 + cosx)/(1 + cosx),

→ cosecx/(1 - cosx) × (1 + cosx)/(1 + cosx)

→ cosecx(1 + cosx)/(1 - cos²x)

→ (1 + cosx)/sinx × 1/(sin²x)

Since we know that, 1 - cos²x = sin²x

→ (1 + cosx)/sin³x = L.H.S

Hence Proved

Answer 2

We are familiar with the Pythagorean trigonometric identity:

$\sin^2x+\cos^2x=1$

On subtracting  $\cos^2x$  from both sides, we get,

$\sin^2x=1-\cos^2x$

The RHS can be factorized.

$\sin^2x=(1+\cos x)(1-\cos x)$

And the LHS can be taken as,

$\dfrac{\sin^3x}{\sin x}=(1+\cos x)(1-\cos x)\\ \\ \\ \Longrightarrow\ \sin^3x\cdot\csc x=(1+\cos x)(1-\cos x)\ \ \ \ \ \ \ \ \ \ \left[\because\ \dfrac{1}{\sin x}=\csc x\right]$

Divide both sides by  $\sin^3x\cdot(1-\cos x).$

$\dfrac{\sin^3x\cdot\csc x}{\sin^3x(1-\cos x)}=\dfrac{(1+\cos x)(1-\cos x)}{\sin^3x(1-\cos x)}\\ \\ \\ \Longrightarrow\ \dfrac{\csc x}{1-\cos x}=\dfrac{1+\cos x}{\sin^3x}\\ \\ \\ \Longrightarrow\ \dfrac{\cos x+1}{\sin^3x}=\dfrac{\csc x}{1-\cos x}$

Thus we got our equation.

Or a general proof is given below.

$\begin{aligned}&\text{LHS}\\ \\ \Longrightarrow\ \ &\dfrac{\cos x+1}{\sin^3x}\\ \\ \Longrightarrow\ \ &\dfrac{(\cos x+1)\frac{1-\cos x}{1-\cos x}}{\sin^3x}\\ \\ \Longrightarrow\ \ &\dfrac{\frac{(\cos x+1)(1-\cos x)}{1-\cos x}}{\sin^3x}\\ \\ \Longrightarrow\ \ &\dfrac{\frac{1-\cos^2x}{1-\cos x}}{\sin^3x}\\ \\ \Longrightarrow\ \ &\dfrac{\frac{\sin^2x}{1-\cos x}}{\sin^2x\cdot\sin x}\\ \\ \Longrightarrow\ \ &\dfrac{\sin^2x}{1-\cos x}\cdot\dfrac{1}{\sin^2x\cdot\sin x}\end{aligned}$

$\Longrightarrow\ \ \dfrac{1}{1-\cos x}\cdot\dfrac{1}{\sin x}\\ \\ \\ \Longrightarrow\ \ &\dfrac{1}{1-\cos x}\cdot\csc x\\ \\ \\ \Longrightarrow\ \ &\dfrac{\csc x}{1-\cos x}\\ \\ \\ \Longrightarrow\ \ \text{RHS}$

Hence proved!