Question

If $\alpha \mathrm{and}\beta$ are the complementary angles, then what is $\sqrt{\mathrm{cosec}\alpha ·\mathrm{cosec}\beta }{\left(\frac{\sin \alpha }{\sin \beta }+\frac{\cos \alpha }{\cos \beta }\right)}^{-1/2}$ equal to:

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Solving $\sqrt{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{}\mathbf{\alpha }\mathbf{·}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{}\mathbf{\beta }}{\left(\frac{sin\alpha }{sin\beta }+\frac{cos\alpha }{cos\beta }\right)}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}$:

It is given that $\alpha \mathrm{and}\beta$ are the complementary angles,

$\Rightarrow \alpha +\beta =90°...\left(1\right)$

$\begin{array}{rcl}\sqrt{\mathrm{cosec}\alpha ·\mathrm{cosec}\beta }{\left(\frac{\sin \alpha }{\sin \beta }+\frac{\cos \alpha }{\cos \beta }\right)}^{-1/2}& =& \sqrt{\frac{1}{\sin \alpha }·\frac{1}{\sin \beta }}{\left(\frac{\sin \alpha }{\sin \beta }+\frac{\cos \alpha }{\cos \beta }\right)}^{-1/2}\\ & =& \frac{1}{{\left(\sin \alpha ·\sin \beta \right)}^{1/2}}{\left(\frac{\left(\sin \alpha ·\cos \beta \right)+\left(\cos \alpha ·\sin \beta \right)}{\sin \beta ·\cos \beta }\right)}^{-1/2}\end{array}$

As $\begin{array}{rcl}\sin (\alpha +\beta )& =& \sin \left(\alpha \right).\cos \left(\beta \right)+\cos \left(\alpha \right).\sin \left(\beta \right)\end{array}$, so

$\begin{array}{rcl}\sqrt{\mathrm{cosec}\alpha ·\mathrm{cosec}\beta }{\left(\frac{\sin \alpha }{\sin \beta }+\frac{\cos \alpha }{\cos \beta }\right)}^{-1/2}& =& \frac{1}{{\left(\sin \alpha ·\sin \beta \right)}^{1/2}}{\left(\frac{\sin (\alpha +\beta )}{\sin \beta ·\cos \left(90°-\alpha \right)}\right)}^{-1/2}\\ & =& \frac{1}{{\left(\sin \alpha ·\sin \beta \right)}^{1/2}}{\left(\frac{\sin 90°}{\sin \beta ·\cos \left(90°-\alpha \right)}\right)}^{-1/2}\left[\mathrm{from}\left(1\right)\right]\\ & =& \frac{1}{{\left(\sin \alpha ·\sin \beta \right)}^{1/2}}{\left(\frac{1}{\sin \beta ·\sin \alpha }\right)}^{-1/2}\left[\mathrm{By}\cos (90°–\theta )=\sin \theta \right]\\ & =& \frac{1}{{\left(\sin \alpha ·\sin \beta \right)}^{1/2}}\times {\left(\sin \alpha ·\sin \beta \right)}^{1/2}\\ & =& 1\end{array}$

Hence, the correct answer is option B.