Question

cos 90 minus theta upon tan theta + cosec 90 - theta into sin theta upon tan 90 minus theta is equal to sec square theta ​

Answer 1

I think this should be the correct answer

Answer 2

Answer:

We need to show that:-

$\frac{\cot(90-\theta)}{\tan \theta}+\frac{cosec (90- \theta) \sin \theta}{\tan (90 -\theta)}= \sec^{2} \theta$

Left hand side

$\frac{\cot (90-\theta)}{\tan \theta}+\frac{cosec (90- \theta) \sin \theta}{\tan (90 -\theta)}$

Since, $cosec (90- \theta)= \sec \theta$ and  $\tan(90- \theta)= \cot \theta$

$\frac{\cot(90-\theta)}{\cot(90 - \theta)}+\frac{\sec \theta \sin \theta}{\cot \theta}$

$1+\frac{\sec \theta \sin \theta}{\cot \theta}$

Since, $\sec \theta =\frac{1}{\cos \theta}$ and $\cot \theta =\frac{\cos \theta}{\sin \theta}$

$1+\frac{\sin \theta}{\cos \theta}{\frac{\cos \theta}{\sin \theta}}$

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

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

$1+(\tan \theta})^{2}$

$1+\tan^{2} \theta$

Using identity $1+\tan^{2} \theta =\sec^{2} \theta$

$\sec^{2} \theta$

= Right hand side

Hence proved