Question

if 1-sin2a/cos2a=tan theta then the value of theta is​

Answer 1

$\textbf{Given:}$

$\mathsf{\dfrac{1-sin2A}{cos2A}=tan\theta}$

$\textbf{To find:}$

$\textsf{The value of}\;\mathsf{\theta}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\dfrac{1-sin2A}{cos2A}=tan\theta}$

$\mathsf{\dfrac{cos^2A+sin^2A-2\,sinA\,cosA}{cos^2A-sin^2A}=tan\theta}$

$\mathsf{\dfrac{(cos-sinA)^2}{cos^2A-sin^2A}=tan\theta}$

$\mathsf{\dfrac{(cosA-sinA)^2}{(cosA-sinA)(cosA+sinA)}=tan\theta}$

$\mathsf{\dfrac{cosA-sinA}{cosA+sinA}=tan\theta}$

$\mathsf{\dfrac{cosA(1-\dfrac{sinA}{cosA})}{cosA(1+\dfrac{sinA}{cosA})}=tan\theta}$

$\mathsf{\dfrac{1-tanA}{1+tanA}=tan\theta}$

$\mathsf{\dfrac{tan\frac{\pi}{4}-tanA}{1+tan\frac{\pi}{4}\,tanA}=tan\theta}$

$\mathsf{tan(\frac{\pi}{4}-A)=tan\theta}$

$\implies\boxed{\mathsf{\theta=\frac{\pi}{4}-A}}$

Answer 2

Given :   (1 - sin2a)/cos2a  = tanθ

To Find :  Value of θ

Solution:

(1 - sin2a)/cos2a  = tanθ

1 = sin²a + cos²a

cos2a = cos²a - sin²a

sin2a = 2sinacosa

=> ( sin²a + cos²a  - 2sinacosa)/( cos²a - sin²a)  =  tanθ

=> ( ±(cosa - sina))²/(cos + sina)(cosa - sina)  =  tanθ

=> ( ±(cosa - sina)) /(cos + sina) = tanθ

Dividing numerator & denominator by cosa in LHS

=> ± ( 1 - tana)/(1 + tana)  = tanθ

1 =  tan(π/4)

=> ±( tan(π/4) - tana)/(1 + tan(π/4).tana)  = tanθ

=> ±tan((π/4) - a)  =  tanθ

=> θ =  (π/4) - a  ,  a -  (π/4)

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