Question

Prove that { cot2 A ( sec A -1) } / ( 1 + sin A) = sec2 A { ( 1 -sin A ) / ( 1 + sec A)}




Pls help at the earliest

Answer 1

$\textbf{To prove:}$

$\mathsf{cot^2A\left(\dfrac{secA-1}{1+sinA}\right)=sec^2A\left(\dfrac{1-sinA}{1+secA}\right)}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{cot^2A\left(\dfrac{secA-1}{1+sinA}\right)}$

$\mathsf{=(cosec^2A-1)\left(\dfrac{secA-1}{1+sinA}\right)}$

$\mathsf{=\left(\dfrac{1}{sin^2A}-1\right)\left(\dfrac{secA-1}{1+sinA}\right)}$

$\mathsf{=\left(\dfrac{1-sin^2A}{sin^2A}\right)\left(\dfrac{secA-1}{1+sinA}\right)}$

$\mathsf{=\left(\dfrac{1^2-sin^2A}{1-cos^2A}\right)\left(\dfrac{secA-1}{1+sinA}\right)}$

$\mathsf{=\left(\dfrac{(1-sinA)(1+sinA)}{1-\dfrac{1}{sec^2A}}\right)\left(\dfrac{secA-1}{1+sinA}\right)}$

$\mathsf{=\left(\dfrac{1-sinA}{\dfrac{sec^2A-1}{sec^2A}}\right)(secA-1)}$

$\mathsf{=sec^2A\left(\dfrac{1-sinA}{sec^2A-1^2}\right)(secA-1)}$

$\mathsf{=sec^2A\left(\dfrac{1-sinA}{secA+1}\right)}$

$\implies\boxed{\mathsf{cot^2A\left(\dfrac{secA-1}{1+sinA}\right)=sec^2A\left(\dfrac{1-sinA}{1+secA}\right)}}$

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Answer 2

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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