Question

Sin theta by cot theta + cosec theta minus sin theta by cos theta minus cosec theta is equals to 2 prove that

Answer 1

$\frac{\text{sin}\theta}{\text{cot}\theta + \text{cosec}\theta} - \frac{\text{sin}\theta}{\text{cot}\theta - \text{cosec}\theta} = 2$ is proved

Step-by-step explanation:

To Prove:

$\frac{\text{sin}\theta}{\text{cot}\theta + \text{cosec}\theta} - \frac{\text{sin}\theta}{\text{cot}\theta - \text{cosec}\theta} = 2$

Left Hand Side:

$\text{sin}\theta \frac{ \text{cot}\theta - \text{cosec}\theta - \text{cot}\theta - \text{cosec}\theta}{(\text{cot}\theta - \text{cosec}\theta) (\text{cot}\theta - \text{cosec}\theta)}$

We know that (a+b)(a-b)= a²-b²

$= \text{sin}\theta \frac{ -2\text{cosec}\theta}{(\text{cot}^2\theta - \text{cosec}^2\theta)}$

$= \frac{ -2\text{sin}\theta \text{cosec}\theta}{(\text{cot}^2\theta - \text{cosec}^2\theta)} -----> (1)$

As,

$\text{sin}\theta = \frac{1}{\text{cosec}\theta}$

$\text{sin}\theta \times \text{cosec}\theta = 1 ----->(2)$

$\text{cot}^2 \theta + 1 = \text{cosec}^2\theta$

$\text{cot}^2 \theta - \text{cosec}^2\theta = 1 ----->(3)$

Substitute the equation 2 and 3 in equation 1,

$\frac{2 \times 1}{1} = 2$

= Right Hand side

LHS = RHS

Hence the given condition is proved.

To Learn More:

1. Sintheta -cos theta /sin theta +cos theta +sin theta +cos theta /sin theta -cos theta =2/2sin²theta -1

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2. Sin theta _ 2 sin3 theta / 2cos3 theta _ cos theta = tan theta

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