Question

sec^2 10 -cot^2 80+sin15 cos75+cos15 sin75÷costita sin (90- tita)+ sin tita cos (90-tita)
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Answer 1

Answer:

3

Step-by-step explanation:

taking thita as A

=( sec^2 10- cot^2 80)+(sin15.cos75) + (cos15.sin75)

+cosA sin(90-A) + sinA cos(90-A)

=(sec^2 10 - tan^2(90-80)) +(sin15.sin(90-75)) +(cos15.cos(90-15)) +cosA.cosA + sinA.sinA

= (sec^2 10 -tan^2 10) +(sin^2 15+ cos^2 15)+ (cos^2 A +sin^2A)

using identities,

sec^2A = tan^2 A +1

and

sin^2 A + cos^2 A=1

= 1+1+1

=3