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Trigonometrical Equations
medium
The number of real numbers $\lambda$ for which the equality $\frac{\sin (\lambda \alpha) \quad \cos (\lambda \alpha)}{\sin \alpha}=\lambda-1$,holds for all real $\alpha$ which are not integral multiples of $\pi / 2$ is
A
$1$
B
$2$
C
$3$
D
infinite
(KVPY-2015)
Solution
(b)
We have,
$\frac{\sin (\lambda, \alpha) \cos (\lambda \alpha)}{\sin \alpha \quad \cos \alpha}=\lambda-1$
$\Rightarrow \cos \alpha \sin \lambda \alpha-\cos (\lambda \alpha)$
$\Rightarrow \sin (\lambda-1) \alpha=\frac{\lambda-1}{2} \sin 2 \alpha$
$\therefore \quad \lambda-1=2 \text { or } \lambda-1=0$
$\therefore \quad \lambda=3 \text { or } \lambda=1$
Hence, $\lambda$ has two values.
Standard 11
Mathematics
