Question

Prove that cot (90 - a)/tan a + cosec (90 - a)/tan (90 - a) * sin a = sec^2 a

Answer 1

Solution:

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i)cot(90-A) = tanA

ii)cosecA(90-A) = secA

iii)tan(90-A) = cotA

iv) secA=1/cosA

v) cotA = cosA/sinA

vi) cos²A+sin²A = 1

____________________

Now,

LHS =cot (90 - a)/tan a + cosec (90 - a)/tan (90 - a) * sin a

=(tanA/tanA)+(secA/cotA)sinA

= 1+ (sin²A/cos²A)

= (cos²A+sin²A)/cos²A

= 1/cos²A

= sec²A

= RHS

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