First slide

Trigonometric Identities

Question

$a \sin x = b \cos x = \frac{2 c \tan x}{1 - \tan^{2} x}$ and ${(a^{2} - b^{2})}^{2} = k c^{2} (a^{2} + b^{2}) \Rightarrow k =$

Moderate

A

1
B

3
C

4
D

2

Solution

$\begin{array}{l} T a k e a \sin x = b \cos x = c \tan 2 x = l (s a y) \\ \Rightarrow \sin x = \frac{l}{a}, \cos x = \frac{l}{b}, \tan 2 x = \frac{l}{c} \end{array} N o w 1 = \sin^{2} x + \cos^{2} x = {(\frac{l}{a})}^{2} + {(\frac{l}{b})}^{2} = l^{2} (\frac{a^{2} + b^{2}}{a^{2} b^{2}})$
$= c^{2} \tan^{2} 2 x (\frac{a^{2} + b^{2}}{a^{2} b^{2}}) = c^{2} {(\frac{2 \tan x}{1 - \tan^{2} x})}^{2} (\frac{a^{2} + b^{2}}{a^{2} b^{2}}) = c^{2} {(\frac{2 \frac{b}{a}}{1 - {(\frac{b}{a})}^{2}})}^{2} (\frac{a^{2} + b^{2}}{a^{2} b^{2}}) (∵ \tan x = \frac{b}{a})$
$\Rightarrow {(a^{2} - b^{2})}^{2} = 4 c^{2} (a^{2} + b^{2}) \Rightarrow k = 4$

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