Question

7. Prove that :
cot2 A (sec A-1)/
1 + sin A=
sec2A
1-sin A\
1 + sec A​

Answer 1

Question :- Prove that {cot²A(sec A -1)} / ( 1 + sin A) = sec² A {( 1 -sin A) / (1 + sec A)}

Solution :-

solving LHS,

→ {cot²A(sec A -1)} / ( 1 + sin A)

Multiply and divide by {(1 - sin A) / (1 - sin A) = 1} and {(sec A + 1) /(sec A + 1)}

→ [{cot²A(sec A -1)} / ( 1 + sin A)] * {(1 - sin A) / (1 - sin A)} * {(sec A + 1) /(sec A + 1)}

Solving Numerator now,

cot²A * (sec A -1) * (1 - sin A) * (sec A + 1)

cot²A * (sec A -1) * (sec A + 1) * (1 - sin A)

using (a + b)(a - b) = a² - b²

cot²A * (sec²A - 1) * (1 - sin A)

using (sec²A - 1) = tan²A

cot²A * tan²A * (1 - sin A)

we know that, cot²A = (1/tan²A)

(1/tan²A) * tan²A * (1 - sin A)

(1 - sin A)

Solving Denominator now,

(1 + sinA) * (1 - sinA) * (sec A + 1)

using (a + b)(a - b) = a² - b²

(1 - sin²A) * (sec A + 1)

using (1 - sin²A) = cos²A

cos²A(sec A + 1)

cos²A(1 + sec A)

using cos²A = (1/sec²A) now,

(1/sec²A)(1 + secA)

therefore, we get,

→ (1 - sinA) /{(1/sec²A)(1 + secA)}

→ sec²A(1 - sin A) / (1 + sec A) = RHS . (Proved.)

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