Question

Prove that tan square theta +cot square theta +2=cosec square theta *sec square theta

Answer 1

i think that u have made a little mistake in wrting the question.... the question should be right is this way

prove that tan2 theta+cot2 theta + 1 = 2 cosec2 theta* sec square theta?
Now by solving according to right ques:
Taking L.H.S:
as tan theta = sin theta divided by cos theta and cot theta is inverse of tan theta

by breaking tan square and cot square theta into their respective formulas
sin2 theta divided by cos2 theta + cos2 theta divided by sin2 theta
Now by taking L.C.M 

sin2 theta + cos2 theta  +1              1+1           as sin2 theta + cos2 theta= 1
-----------------------------          =    ---------------
(cos2 theta) (sin2 theta)             (cos2 theta) ( sin2theta)

       2
 ----- ---------   
 (cos2 theta) ( sin2teta)

as sec2theta = 1                       and              cosec2theta = 1            
                        ------                                                        ----------
                         cos2theta                                                sin2theta

so 2  ( 1 divided by cos2theta) = 2 sec square  theta
and 2  ( 1 divided by sin2 theta ) = 2 cosec square theta


L.H.S = R.H.S 
Hence prooved...

Answer 2

Given:

$tan^{2}\alpha +cot^{2}+2=cosec^{2}\alpha .sec^{2}\alpha$

To Find:

We need to prove the given equation.

Solution:

we should be aware of trigonometric identities to solve this problem.

The identities used in this problem are

$sec^{2}\alpha -tan^{2}\alpha =1\\cosec^{2}\alpha -cot^{2}\alpha =1$

$sin^{2}\alpha +cos^{2}\alpha =1$

from the above $tan^{2}\alpha =sec^{2}\alpha -1$

$cot^{2}\alpha =cosec^{2}\alpha -1$

substitute these values in the equation. Consider LHS,

$=sec^{2}\alpha -1+cosec^{ 2}\alpha -1+2$

$=sec^{2}\alpha +cosec^{2}\alpha$

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

$=\frac{sin^{2}\alpha +cos^{2}\alpha }{sin^{2}\alpha .cos^{2}\alpha }$

$=\frac{1}{sin^{2}\alpha .cos^{2}\alpha }$

$=cosec^{2}\alpha .sec^{2} \alpha$

LHS=RHS

Therefore, we have proved that $tan^{2}\alpha +cot^{2}+2=cosec^{2}\alpha .sec^{2}\alpha$.

#SPJ2