If tan-n sin-cos-1-n sin 2- then tan - is equal to

The correct option is C tan β=n ta α/1+tan^2α + n tan^2 α=n tan α/1+(1 n)tan^2 α=tan(α β)=tan αn tan α/1+(1 n)tan^2 α/1+tan α, n tan α/1+(1 n)tan^2α =tan α ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

If tanβ=n sin...

Question

$I f t a n β = \frac{n s i n α c o s α}{1 - n s i n^{2} α}, t h e n t a n (α - β) i s e q u a l t o$

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Solution

The correct option is B

$t a n β = \frac{n t a α}{1 + t a n^{2} α + n t a n^{2} α} = \frac{n t a n α}{1 + (1 - n) t a n^{2} α} = t a n (α - β) = \frac{t a n α \frac{n t a n α}{1 + (1 - n) t a n^{2} α}}{1 + \frac{t a n α, n t a n α}{1 + (1 - n) t a n^{2} α}} = \frac{t a n α (1 - n) t a n^{3} α - n t a n α}{1 + (1 - n) t a n^{2} α + n t a n^{2} α} = \frac{(1 - n) t a n α (1 + t a n^{2} α)}{1 + t a n^{2} α} = (1 - n) t a n α$

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Similar questions

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If tan β=n sin α cos α1−n sin2 α, then tan (α−β)is equal to

The correct option is B tan β=n ta α1+tan2α+n tan2 α=n tan α1+(1−n)tan2α=tan(α−β)=tan αn tan α1+(1−n)tan2α1+tan α, n tan α1+(1−n)tan2α=tan α(1−n)tan3α−n tan α1+(1−n)tan2α+n tan2α=(1−n)tan α(1+tan2α)1+tan2α=(1−n)tan α

$I f t a n β = \frac{n s i n α c o s α}{1 - n s i n^{2} α}, t h e n t a n (α - β) i s e q u a l t o$