Question

costheta/1-sin theta+1-sintheta/costheta=2sectheta​

Answer 1

Answer:

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Answer 2

$\frac{\text{cos}\theta}{1- \text{sin}\theta} + \frac{1- \text{sin}\theta}{\text{cos}\theta} = 2\text{sec}\theta$

Explanation:

To prove :

$\frac{\text{cos}\theta}{1- \text{sin}\theta} + \frac{1- \text{sin}\theta}{\text{cos}\theta} = 2\text{sec}\theta$

Left hand side-

$\Rightarrow \frac{\text{cos}\theta}{1- \text{sin}\theta} + \frac{1- \text{sin}\theta}{\text{cos}\theta}$

$\Rightarrow \frac{\text{cos}\theta \times \text{cos}\theta+ (1- \text{sin}\theta) \times (1- \text{sin}\theta)}{(1- \text{sin}\theta) \text{cos}\theta}$

$\Rightarrow \frac{\text{cos}^2\theta + (1- \text{sin}\theta)^2}{(1- \text{sin}\theta) \text{cos}\theta}$

$\Rightarrow \frac{\text{cos}^2\theta + (1- 2\text{sin}\theta +\text{sin}^2\theta)}{(1- \text{sin}\theta) \text{cos}\theta}$

$\Rightarrow \frac{\text{cos}^2\theta + 1- 2\text{sin}\theta +\text{sin}^2\theta}{(1- \text{sin}\theta) \text{cos}\theta}$

We know that,

$\text{sin}^2\theta + \text{cos}^2\theta = 1$

$\Rightarrow \frac{1 + 1- 2\text{sin}\theta}{(1- \text{sin}\theta) \text{cos}\theta}$

$\Rightarrow \frac{2 - 2\text{sin}\theta}{(1- \text{sin}\theta) \text{cos}\theta}$

$\Rightarrow 2 \times \frac{(1- \text{sin}\theta)}{(1- \text{sin}\theta) \text{cos}\theta}$

We can cancel 1-sin theta on Numerator and in Denominator

$\Rightarrow 2 \times \frac{1}{\text{cos}\theta}$

$\text{Where} \frac{1}{\text{cos}\theta} = \text{sec}\theta$

$\Rightarrow 2 \ \text{sec}\theta$

= Right Hand side

Hence Right Hand Side = Left Hand Side

$\frac{\text{cos}\theta}{1- \text{sin}\theta} + \frac{1- \text{sin}\theta}{\text{cos}\theta} = 2\text{sec}\theta$

Therefore, the condition is satisfied.

To Learn More:

1.Prove that 1+sin theta/cos theta +cos theta/1+sin theta=2sec theta

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2. Sin 90-theta by 1+sin theta +cos theta by 1-cos 90 theta =2sec theta

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