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Question
Prove the following:
`1/(1 + sinA) - 1/(1 - sinA) = - 2secA tanA`
Theorem
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Solution
LHS = `((1 - sinA) - (1 + sinA))/((1 + sinA)(1 - sinA))`
`(1 - sinA - 1 - sinA)/((1 + sinA)(1 - sinA))`
= `(-2sinA)/((1 + sinA)(1 - sinA))`
Simplify the denominator:
`(-2sinA)/(1 - sin^2A)`
1 − sin2 A = cos2 A
∴ `(-2sinA)/(cos^2A)`
Separate the terms to match the RHS = cos2 A as cos A.cos A
`-2. 1/cosA.sinA/cosA`
Substitute the trigonometric definitions:
`1/cosA = secA and sinA/cosA = tanA`
− sec A . tan A
LHS = RHS
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