Question

prove that cosec^2theta +tan^2 (90-theta)=2sec^2(90-theta)-1​

Answer 1

Solution:

$cosec^{2}\theta+tan^{2}(90-\theta) = 2sec^{2}(90-\theta)-1$

Now, let's take

L.H.S = $cosec^{2}\theta+tan^{2}(90-\theta)$

and

R.H.S = $2sec^{2}(90-\theta)-1$

Therefore,

         L.H.S => $\frac{1}{sin^{2}\theta} } + cot^{2}\theta$

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

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

                =>$\frac{1+1-sin^{2}\theta}{sin^{2}\theta}$

               =>$\frac{2-sin^{2}\theta}{sin^{2}\theta}$

     

       R.H.S  =>$2 * (cosec^{2}\theta)-1$

                   => $2*\frac{1}{sin^{2}\theta} -1$

                  => $\frac{2-sin^{2}\theta}{sin^{2}\theta}$

Hence proved!!!

   

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