Question

(tan theta+ cot theta)^2= sec^2 theta+cosec^2 theta​

Answer 1

Answer:

Proved that $(\tan \theta + \cot \theta)^{2} = \sec^{2} \theta + \csc^{2} \theta$.

Step-by-step explanation:

We have to prove that  

$(\tan \theta + \cot \theta)^{2} = \sec^{2} \theta + \csc^{2} \theta$

Now, Left hand side  

= $(\tan \theta + \cot \theta)^{2}$

= $\tan^{2} \theta + 2\times \tan \theta \times \cot \theta + \cot^{2} \theta$ {Since we know the formula (a + b)² = a² + 2ab + b²}

=  $\tan^{2} \theta + 2 + \cot^{2} \theta$

{Since $\tan \theta \times \cot \theta = 1$}

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

= $\sec^{2} \theta + \csc^{2} \theta$

{Since we know the identities:

$\sec^{2} x = \tan^{2} x + 1$ and $\csc^{2} x = \cot^{2} x + 1$}

= Right hand side

Hence, proved.