Question

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-tana.cot(90-a)+seca.cosec(90-a)+sin35+sin55 / tan10.tan20.tan30.tan70.tan80

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Answer 1

Answer:

cot(90-a) = tana

cosec(90-a) = seca

sin35 = sin(90-55) = coe35

tan10 is opposite of tan80

tan 20 is opposite of tan 70

sin²a+cos²a = 1

sec²a -tan²a = 1

tan30 = 1/√3

now their is little bit mistake in writing the question , but wo samaj aa rhi hai, sin²35 and sin²55 hai wo..

Now,

using all these value we get :-

(-tana²a+sec²)+(sin²35+cos²35)/tan10*tan20*tan30*cot10*cot20)

= 1+1/√3

= 2/√3

= 2√3/3 (Ans)

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Answer 2

this will help you

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