If tan -n sin - cos -1-n sin2 - then tan - - - is equal to

We have given tan β=n sin α cos α1 n sin^2 α ⇒ tan β=n sin α cos αsec^2α n tan^2 α ⇒ tan β=n tan α(1+tan^2α) nbsp;n tan^2 α ⇒ tan β=n tan α1+(1 n)tan^2 α ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles: Tangent and Cotangent Functions

If tan β=n si...

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

We have given
$t a n β = \frac{n s i n α c o s α}{1 - n s i n^{2} α}$
$\Rightarrow t a n β = \frac{n \frac{s i n α}{c o s α}}{s e c^{2} α - n t a n^{2} α}$

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

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

Hence option(b) is correct answer.

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

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