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Question
If tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.
Options
a
b
`a/b`
None
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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