If tan-4-y-2-tan 3-4-x-2- then sinx3-sin 2 x-1-3 sin 2 x equals

The correct option is A sin y1+tany/2/1 tany/2= ( 1+tany/2/1 tany/2 )^3 Square both sides, we get 1+siny/1 siny=(1+sinx)^3/(1 sinx)^3 Using componendo and divi ...

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Byju's Answer

Standard XII

Mathematics

Range of a Function

If tanπ/4+y/2...

Question

$I f t a n (\frac{π}{4} + \frac{y}{2}) = t a n^{3} (\frac{π}{4} + \frac{x}{2}), t h e n s i n x (\frac{3 + s i n^{2} x}{1 + 3 s i n^{2} x}) e q u a l s$

cos y

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sin y

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sin 2y

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Solution

The correct option is B sin y
$\frac{1 + t a n \frac{y}{2}}{1 - t a n \frac{y}{2}} = {(\frac{1 + t a n \frac{y}{2}}{1 - t a n \frac{y}{2}})}^{3}$
Square both sides, we get
$\frac{1 + s i n y}{1 - s i n y} = \frac{(1 + s i n x)^{3}}{(1 - s i n x)^{3}}$
Using componendo and dividendo
$\frac{2 s i n y}{2} = \frac{(3 + s i n^{2} x)}{1 + 3 s i n^{2} x} s i n x$

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Similar questions

Q. If

tan (\frac{π}{4} + \frac{y}{2})

{tan}^{3} (\frac{π}{4} + \frac{x}{2})

, prove that -

sin y = sin x . \frac{3 + {sin}^{2} x}{1 + 3 {sin}^{2} x} .

Q. If

t a n (\frac{π}{4} + θ) + t a n (\frac{π}{4} - θ) = a

then

t a n^{3} (\frac{π}{4} + θ) + t a n^{3} (\frac{π}{4} - θ)

Q. If

0 < x < \frac{π}{2}

, then

tan (\frac{π}{4} + \frac{x}{2}) + tan (\frac{π}{4} - \frac{x}{2})

is equal to

Q. If

tan (\frac{π}{4} + θ) \times tan (\frac{3 π}{4} + θ) = m

. Find

- m

Q. Consider the following:
1.

tan (\frac{π}{6})

tan (\frac{3 π}{4})

tan (\frac{5 π}{4})

tan (\frac{2 π}{3})

What is the correct order?

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Iftan(π4+y2)=tan3(π4+x2),thensinx(3+sin2x1+3sin2x)equals

The correct option is B sin y1+tany21−tany2=(1+tany21−tany2)3 Square both sides, we get 1+siny1−siny=(1+sinx)3(1−sinx)3 Using componendo and dividendo 2siny2=(3+sin2x)1+3sin2xsinx

$I f t a n (\frac{π}{4} + \frac{y}{2}) = t a n^{3} (\frac{π}{4} + \frac{x}{2}), t h e n s i n x (\frac{3 + s i n^{2} x}{1 + 3 s i n^{2} x}) e q u a l s$