Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Given that:

$a\sin x=b\cos x=\frac{2c\tan x}{1-{\tan }^{2}x}$

and $(a^2-b^2{)}^2=k{c}^2(a^2+b^2)$

To find:

$k=?$

Solution:

$a\sin x=b\cos x$

or, $\tan x=\frac{b}{a}$

$a\sin x=\frac{2c\tan x}{1-{\tan }^{2}x}$

or, $a\sin x=\frac{2c\times \frac{b}{a}}{1-{\left(\frac{b}{a}\right)}^2}$

or, $a\times \frac{b}{\sqrt{a^2+b^2}}=\frac{2abc}{a^2-b^2}$

or, $\frac{1}{\sqrt{a^2+b^2}}=\frac{2c}{a^2-b^2}$

or, $(a^2-b^2)=2c\sqrt{a^2+b^2}$

Squaring both sides we get,

or, $(a^2-b^2{)}^2=4c^2(a^2+b^2)$

So, value of $k=4$

Therefore, D is the correct option.