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If {^3}{1+^2}= - a . Find a

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If $\frac{tan ^{3} α}{1 + tan ^{2} α} = sec α sin α - a sin α cos α$ . Find $a$

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If

\frac{tan ^{3} α}{1 + tan ^{2} α} = sec α sin α - a sin α cos α

. Find

a

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Text solutionVerified

LHS

\frac{tan ^{3} α}{1 + tan ^{2} α}

= \frac{\frac{sin ^{3} α}{cos ^{3} α}}{1 + \frac{sin ^{2} α}{cos ^{2} α}} = \frac{\frac{sin ^{3} α}{cos ^{3} α}}{\frac{cos ^{2} α + sin ^{2} α}{cos ^{2} α}}

∵ sin^{2} x + cos^{2} x = 1

= \frac{sin ^{3} α cos ^{2} α}{cos ^{3} α} = \frac{sin ^{3} α}{cos α}

= \frac{sin α ( sin ^{2} α )}{cos α}

= \frac{sin α ( 1 - cos ^{2} α )}{cos α}

= \frac{sin α - sin α cos ^{2} α}{cos α}

= \frac{sin α}{cos α} - sin α cos α

= sin α sec α - sin α cos α

\Rightarrow α = 1

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Question Text	If $\frac{tan ^{3} α}{1 + tan ^{2} α} = sec α sin α - a sin α cos α$ . Find $a$
Answer Type	Text solution:1
Upvotes	150