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If tanθ = ab, then bcos2θ + asin2θ is equal to ______.

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Question

If tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.

Options

  • a

  • b

  • `a/b`

  • None

MCQ
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Solution

If tanθ = `a/b`, then bcos2θ + asin2θ is equal to b.

Explanation:

Given that: tanθ = `a/b`

bcos2θ + asin2θ = `b[(1 - tan^2 theta)/(1 + tan^2 theta)] + a[(2 tan theta)/(1 + tan^2 theta)]`

= `b[(1 - a^2/b^2)/(1 + a^2/b^2)] + a[((2a)/b)/(1 + a^2/b^2)]`

= `b[(b^2 - a^2)/(b^2 + a^2)] + [((2a^2)/b)/((b^2 + a^2)/b^2)]`

= `(b^3 - a^2b)/(b^2 + a^2) + (2a^2b)/(b^2 + a^2)`

= `(b^3 - a^2b + 2a^2b)/(b^2 + a^2)`

= `(b^3 + a^2b)/(b^2 + a^2)`

= `(b(b^2 + a^2))/(b^2 + a^2)`

= b

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Chapter 3: Trigonometric Functions - Exercise [Page 59]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 58 | Page 59

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