Question

sec square (90-theta)-cot^2 theta/2(sin square 25+sin square 65)+2 sin square 30 tan square 32 tan square 58/3(sec square 33-cot square 57)

Answer 1

Answer:

2/3

Step-by-step explanation:

Sec²(90 - θ) - Cot²θ/ 2(Sin²25 + Sin²65) + 2Sin²30Tan²32Tan²58/3(Sec²33 - Cot²57)

Remember

Sec (90 - θ) = Cosecθ

Sin 25 = Sin (90-65) = Cos65

Tan 58 = Tan (90 - 32) = Cot32

Sec 33 = Sec (90 - 57) = Cosec 57

Sin 30 = 1/2

So now substitute the above in given equation

= Cosec²θ - Cot²θ / 2(Cos²65 + Sin²65)  

+  2 * (1/2)²Tan²32Cot²32 / 3(Cosec²57 - Cot²57)

Remember Cosec²θ - Cot²θ = 1

                  Tanθ*Cotθ = 1

                  Sin²θ + Cos²θ = 1

=  1/2 + 1/ 6

=  4/6

= 2/3