Question

$\sf\dfrac{2sinx}{sin3x}+ \dfrac{tanx}{tan3x}$

Answer 1

EXPLANATION.

$\sf \implies \dfrac{2Sinx}{Sin 3x} \ + \ \dfrac{tanx}{tan 3x}$

As we know that,

Formula of :

⇒ sin3x = 3sinx - 4sin³x.

⇒ tan3x = 3tanx - tan³x/1 - 3tan²x.

Using this formula in equation, we get.

$\sf \implies \dfrac{2sinx}{3sinx - 4sin^{3}x } \ + \ \dfrac{tanx}{\dfrac{3tanx - tan^{3}x }{1 - 3tan^{2}x } }$

$\sf \implies \dfrac{2sinx}{3sinx - 4sin^{3}x } \ + \ \dfrac{tanx(1 - 3tan^{2}x) }{3tanx - tan^{3} x}$

$\sf \implies \dfrac{2sinx}{sinx(3 - 4sin^{2}x) } \ + \ \dfrac{tanx(1 - 3tan^{2} x)}{tanx(3 - tan^{2}x) }$

$\sf \implies \dfrac{2}{3 - 4sin^{2}x } \ + \ \dfrac{1 - 3tan^{2} x}{3 - tan^{2}x }$

As we know that,

⇒ tan²x = sin²x/cos²x.

Using this formula in equation, we get.

$\sf \implies \dfrac{2}{3 - 4sin^{2}x } \ + \ \dfrac{1 - \dfrac{3sin^{2}x }{cos^{2}x } }{3 - \dfrac{sin^{2} x}{cos^{2}x } }$

$\sf \implies \dfrac{2}{3 - 4sin^{2}x } \ + \ \dfrac{\dfrac{cos^{2}x - 3sin^{2} x }{cos^{2}x } }{\dfrac{3cos^{2} x - sin^{2} x}{cos^{2}x } }$

$\sf \implies \dfrac{2}{3 - 4sin^{2}x } \ + \ \dfrac{cos^{2}x - 3sin^{2} x }{3cos^{2} x - sin^{2} x}$

As we know that,

Formula of :

⇒ cos²x = 1 - sin²x.

Using this formula in equation, we get.

$\sf \implies \dfrac{2}{3 - 4sin^{2}x } \ + \ \dfrac{(1 - sin^{2} x) - 3sin^{2}x }{3( 1 - sin^{2} x) - sin^{2}x }$

$\sf \implies \dfrac{2}{3 - 4sin^{2}x } \ + \ \dfrac{1 - sin^{2}x - 3sin^{2}x }{3 - 3sin^{2}x - sin^{2}x }$

$\sf \implies \dfrac{2}{3 - 4sin^{2}x } \ + \ \dfrac{1 - 4sin^{2} x}{3 - 4sin^{2}x }$

$\sf \implies \dfrac{2 + (1 - 4sin^{2}x) }{3 - 4sin^{2}x }$

$\sf \implies \dfrac{3 - 4sin^{2}x }{3 - 4sin^{2}x } = 1$

$\sf \implies \dfrac{2Sinx}{Sin 3x} \ + \ \dfrac{tanx}{tan 3x} = 1$

                                                                                                                         

MORE INFORMATION.

Trigonometric ratios of multiple angles.

(1) = sin2x = 2sinx.cosx = 2tanx/1 + tan²x.

(2) = cos2x = cos²x - sin²x = 2cos²x - 1 = 1 - 2sin²x = 1 - tan²x/1 + tan²x.

(3) = tan2x = 2tanx/1 - tan²x.

(4) = sin3x = 3sinx - 4sin³x.

(5) = cos3x = 4cos³x - 3cosx.

(6) = tan3x = 3tanx - tan³x/1 - 3tan²x.

(7) = sin(x/2) = √1 - cosx/2.

(8) = cos(x/2) = √1 + cosx/2.

(9) = tan(x/2) = √1 - cosx/1 + cosx = 1 - cosx/sinx = sinx/1 + cosx.

Answer 2

$\colorbox{lightgreen}{☑ \: Verified Answer}$

★ Concept :

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★ Cos function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse.

★ The cosine function is one of the three main primary trigonometric functions and it is itself the complement of sine(co+sine).

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Given :

$\implies \sf \frac{2sinx}{sin3x} + \sf\frac{tanx}{tan3x}$

$\implies \sf \: \frac{2 \: sin \: x}{3 \: \sin x - 4 \sin {}^{3}x } \: + \frac{ \tan x }{ \sf \frac{3 \tan x - \tan {}^{3}x}{1 - 3 \tan {}^{2} x } }$

Now :

$= \sf \: \frac{2 \sin x }{sin \: x (3 - 4 \: sin {}^{2} x)} \: + \frac{tan \: x(1 - 3 \: tan {}^{2} x}{tan \: x(3 - tan {}^{2}x) }$

$\sf \: \frac{2}{3 - 4 \: sin {}^{2}x}\: + 1 - \frac{3 \: sin {}^{2}x }{cos {}^{2} x} / 3 - \frac{sin {}^{2}x}{cos {}^{2}x}$

$\implies \sf \: \frac{2}{3 - 4 \: sin {}^{2}x} \: + \frac{cos {}^{2}x - 3 \: sin {}^{2}x}{3 \: cos {}^{2}x - sin {}^{2}x}$

$\implies \sf \: \frac{2}{3 - 4 \: sin {}^{2}x} \: + \frac{1 - sin {}^{2}x - 3 \: sin {}^{2}x}{3 - 3 \: sin {}^{2}x - sin {}^{2}x}$

$\implies \sf \: \frac{2}{3 - 4 \: sin {}^{2}x} + \frac{1 - 4 \: sin {}^{2}x}{3 - 4 \: sin {}^{2}x}$

$\implies \sf \: \frac{2 + 1 - 4 \: sin {}^{2}x}{3 - 4 \: sin {}^{2}x}$

Finally :

$\implies \sf \: \frac{3 - 4 \: sin {}^{2}x}{3 - 4 \: sin {}^{2}x} = 1$

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★ Be Brainly :)

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