Question

tan theta/ tan (90°- theta) + sin (90° - theta)/cos theta= sec2 theta

Answer 1

Tan theta/cot theta + cos theta/cos theta
=>sin2 theta/cos2 theta + 1
=>sin2 theta + cos2 theta/cos2 theta
=>1/cos2 theta
=>sec2 theta
Hence proved

Answer 2

To prove that: $\dfrac{\tan \theta}{\tan (90- \theta)} + \dfrac{\sin (90 - \theta)}{\cos \theta} =\sec^2 \theta.$

Solution:

L.H.S. = $\dfrac{\tan \theta}{\tan (90- \theta)} + \dfrac{\sin (90 - \theta)}{\cos \theta}$

Using the trigonometric identity:

$\cot \theta$ = $\tan (90- \theta)$ and

$\cos \theta$ = $\sin (90- \theta)$

= $\dfrac{\tan \theta}{\cot \theta} + \dfrac{\cos \theta}{\cos \theta}$

= $\dfrac{\tan \theta}{\cot \theta} + 1$

Using the trigonometric identity:

$\tan \theta$ = $\dfrac{1}{\cot \theta}$

= $\dfrac{\tan \theta}{\dfrac{1}{\tan \theta} } + 1$

= $\tan^2 \theta$ + 1

Using the trigonometric identity:

$\sec^2 \theta$ - $\tan^2 \theta$ = 1

⇒ $\sec^2 \theta$ = 1 + $\tan^2 \theta$

= $\sec^2 \theta$

= R.H.S., proved.

Thus, $\dfrac{\tan \theta}{\tan (90- \theta)} + \dfrac{\sin (90 - \theta)}{\cos \theta} =\sec^2 \theta$, proved.