Question

If (1+sin2x)/(1-sin2x) = cot²(a+x) ∀ x ∈ R ~ (nπ + π/4), n ∈ N, then a can be

(a) π/4

(b) π/2

(c) 3π/4

(d) None of these

Answer 1

Question:

If (1+sin2x)/(1-sin2x) = cot²(a+x)

∀ x ∈ R ~ (nπ + π/4), n ∈ N, then a can be :

Formula's used:

$\sf1) \tan(x + y) = \dfrac{ \tan x + \tan y}{1 - \tan x \times \tan y}$

$\sf 2)\tan(x - y) = \dfrac{ \tan x - \tan y}{1 +\tan x \times \tan y}$

$\sf3)\sin{}^{2}x+\cos{}^{2}x=1$

$\sf4)\sin2x=2\sin\:x\cos\:x$

Solution :

$\sf \dfrac{1 + \sin2x}{1 - \sin2x} = \cot {}^{2} (a + x)$

First solve

LHS $= \sf \dfrac{1 + \sin2x}{1 - \sin2x}$

$\sf{=\frac{1+2\:sin\:x\:cos\:x}{1-2\:sin\:x\:cos\:x}}$

$\sf = \dfrac{ \sin {}^{2} x + \cos {}^{2} x + 2 \sin x \cos x}{ \sin {}^{2}x + \cos {}^{2} x - 2 \sin x \cos x }$

$\sf = \dfrac{( \cos x + \sin x) {}^{2} }{( \cos x - \sin x) {}^{2} }$

Now divide both Numerator and denominator by cos²x

$\sf = \dfrac{\frac{(\cos x + \sin x) {}^{2} }{ \cos {}^{2} x} }{ \frac{ ( \cos x - \sin x) {}^{2} }{ \cos {}^{2} x} }$

$\sf = \dfrac{(1 + \tan x) {}^{2} }{(1 - \tan x) {}^{2} }$

$\sf = ( \dfrac{1}{ \frac{1 - \tan x}{1 + \tan x} } ) {}^{2}$

$\sf = ( \dfrac{1}{ \frac{ - ( \tan x - 1)}{1 + \tan x} } ) {}^{2}$

$\sf = ( \dfrac{1}{ \frac{ \tan x - 1}{1 + \tan x} }) {}^{2}$

We know that tan 3π/4 = -1

$\sf = (\dfrac{1}{ \frac{ \tan x + \ \tan \frac{3\pi}{4} }{1 - \tan \frac{3\pi}{4} \times \tan x } } ) {}^{2}$

We know that tan(A+B) =( tanA+tanB)/1-tanAtanB

$\sf = ( \dfrac{1}{ \tan( \frac{3\pi}{4} + x)}) {}^{2}$

$\sf = \cot {}^{2} ( \frac{3\pi}{4} + x)$

RHS $\sf = \cot {}^{2} (a + x)$

On comparing LHS and RHS

$\sf \cot {}^{2} ( \frac{3\pi}{4} + x) = \cot {}^{2} (a + x)$

$\sf \implies a = \frac{3\pi}{4}$

Correct option c) 3π/4

Answer 2

Given that :-

If (1+sin2x)/(1-sin2x) = cot²(a+x) ∀ x ∈ R ~ (nπ + π/4), n ∈ N, then a can be

Solution :-

Firstly we'll solve the LHS .

LHS = $\sf{ \frac{1 + \sin 2x}{1 - \sin 2x}}$

We know that sin²x + cos²x = 1 and 2sinxcox = sin2x

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

Now we got the form a²+b²+2ab = (a+b)²

$\sf{\implies {\sin}^{2}x + {\cos}^{2}x + 2 \sin x \cos x = {(\cos x + \sin x )}^{2}}$

$\sf{\implies {\sin}^{2}x + {\cos}^{2}x - 2 \sin x \cos x = {(\cos x - \sin x )}^{2}}$

Now writing it in fraction again .

$\sf{\implies \frac{{(\cos x + \sin x }^{2}}{{\cos x - \sin x}^{2} }}$

Now multiplying and dividing by cos² x .

$\sf{\implies \frac{{(\cos x + \sin x)}^{2}}{{(\cos x - \sin x )}^{2}} \times \frac{{\cos x}^{2} }{{\cos x}^{2}} }$

→ (cosx + sinx/cox)²/ (cosx-sinx/cosx)²

$\sf{\implies ({\frac{1 + \tan x }{1 - \tan x}}^{2}})$

$\sf{\implies ({\frac{\frac{1}{1- \tan x }}{1 + \tan x}})^{2}}$

$\sf{\implies ({\frac{\frac{1}{\tan x - 1}}{1 + \tan x}})^{2}}$

$\sf{\implies ({\frac{\frac{1}{ \tan x - 1}}{1 - (-1) \tan x}})^{2}}$

As we know that :-

→ tan 135° = tan ( 180-45°)

→ - tan(45°) = - 1 .

So replacing -1 by tan 135° or tan 3π/4

$\sf{\implies ({\frac{\frac{1}{ \tan x - \tan \frac{3 \pi}{4}}}{1 - \tan \frac{3 \pi}{4} \tan x}})^{2}}$

→ tan(a+b) = tana + tanb/1 - tana . tanb

$\sf{\implies ({\frac{1}{\tan( \frac{3 \pi }{4} + x}})^{2}}$

→ $\sf{\implies \frac{1}{\tan x} = \cot x }$

→ $\sf{\implies( {\cot \frac{3 \pi}{4} + x})^{2}}$ = LHS .

Now comparing LHS and RHS .

$\sf{\implies ({\cot \frac{3 \pi}{4} + x})^{2}}$ = $\sf ({\cot a + x})^{2}$

$\boxed{\sf{a = \frac{3 \pi}{4}}}$