Question

solve it now guys please ​
and never judge the handwriting

Answer 1

Step-by-step explanation:

your answer attached in the photo

Answer 2

We need to prove

$(\frac{1}{sec^2 theta - cos^2 theta} ) + (\frac{1}{ cosec^2 theta -  sin^2 theta}) ^{sin^2 theta cos^2 theta} = 1 - sin^2 theta cos^2 theta$

divided by ( 2 + sin² theta cos² theta )

So L.H.S = $(\frac{sin^2 theta cos^2 theta}{sec^2 theta - cos^2 theta} )$ + $(\frac{sin^2 theta cos^2 theta}{ cosec^2 theta - sin^2 theta} )$

= $(\frac{ sin^2 theta cos^4 theta}{ 1 - cos^4 theta}) + (\frac{sin^4 theta cos^2 theta}{ 1 - sin^4 theta} )$

Now break ( 1 – $cos^{4}  theta$ ) into ( 1 + cos² theta )( 1 – cos² theta) and ( 1 – cos² theta)= sin² theta so it can be terminated from numerator and denominator, do the same for the one with ( 1 – $sin^{4}$ theta) in denominator

so now we get

L.H.S = $(\frac{cos^4 theta}{1 + cos^2 theta} ) + (\frac{sin^4 theta}{1 + sin^2 theta} )$

= $(\frac{cos^4 theta + sin^4 theta + sin^2 theta cos^4 theta + sin^4 theta cos^2 theta}{1 + sin^2 theta + cos^2 theta + sin^2 theta cos^2 theta} )$

Now $sin^{4} theta + cos^{4} theta = 1 - 2sin^2 theta cos^2 theta$

so L.H.S = $[\frac{1 - 2sin^2 theta cos^2 theta + sin^2 theta cos^2 theta( sin^2 theta + cos^2 theta)}{1 + sin^2 theta + cos^2 theta + sin^2 theta cos^2 theta} ]$

Using the formula sin² A + cos² A = 1

we get

L.H.S= $(\frac{1 - sin^2 theta cos^2 theta}{2 + sin^2 theta cos^2 theta} )$ = R.H.S

Hope That Helped You Titan..

: )