Question

Solve only . 10 .a
(Trigonometric ratios of multiple angles)
#salute U if u solved

Answer 1

$\textbf{To prove:}$

$\dfrac{1-tan\,B}{1+tan\,B}=\dfrac{1-sin\,2B}{cos\,2B}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{1-tan\,B}{1+tan\,B}$

$=\dfrac{1-\frac{sinB}{cosB}}{1+\frac{sinB}{cosB}}$

$=\dfrac{\frac{cosB-sinB}{cosB}}{\frac{cosB+sinB}{cosB}}$

$=\dfrac{cosB-sinB}{cosB+sinB}$

$\text{Multily both numerator and denominator by cosB-sinB ,}$

$\text{we get}$

$=\dfrac{cosB-sinB}{cosB+sinB}{\times}\dfrac{cosB-sinB}{cosB-sinB}$

$=\dfrac{(cosB-sinB)^2}{cos^2B-sin^2B}$

$=\dfrac{cos^2B+sin^2B-2\,sinB\,cosB}{cos^2B-sin^2B}$

$\text{Using the following identities}$

$\boxed{\bf\,sin^2A+cos^2A=1}$

$\boxed{\bf\,sin2A=2\,sinA\,cosA}$

$\boxed{\bf\,cos2A=cos^2A-sin^2A}$

$\text{we get}$

$=\dfrac{1-sin\,2B}{cos\,2B}$

$\implies\boxed{\bf\dfrac{1-tan\,B}{1+tan\,B}=\dfrac{1-sin\,2B}{cos\,2B}}$

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