Question

solve this question.

Answer 1

[tex][sec ^{2} (90-a) - cot^{2} a]/ 2( sin^{2} 65 + sin^{2} 25)

 sec^{2} (90-a)[/tex] can be written as $cosec^{2} a$
so 

$(cosec^{2} a - cot^{2} a)/ 2 ( sin^{2}65 + sin^{2} 25)$

$sin^{2} 25$ can be written as $sin^{2} (90-65)$ which is 

$cos^{2}65$

as $cosec^{2} a - cot^{2} a = 1$

so the equation becomes 

$(cosec^{2} a-cot^{2} a)/ 2 ( sin^{2} 65 + cos^{2} 65)$

$1/ 2 (1) =1/2............(1)$

[tex](2 sin^{2} 30.tan^{2} 32.tan 58^{2} )/3(1) [/tex]

[tex]2 (1/4)(1)/ 3 =\ \textgreater \ 1/6 [/tex].........(2)

$cos ^{2} 30/sin ^{2} 60 =\ \textgreater \ 1..........(3)$


now adding (1),(2) and (3) we get the answer 

(1/2) + (1/6)  + 1

which is equal to 5/3

Answer 2

HELLO DEAR,

we know that:-

sec²(90°-A)=cosec²A

sin²(90-A)=cos²A n

cot²(90-A)=tan²A


I HOPE ITS HELP YOU DEAR,
THANKS