Question

Statement (I): If
$\sin \alpha =\frac{12}{13},(0<\alpha <\frac{\pi }{2})$ and
$\cos \beta =-\frac{3}{5},(\pi <\beta <\frac{3\pi }{2})$ then

$\sin (\alpha +\beta )=\frac{56}{65}$

Statement (II): If

$\theta$ and $\varphi$ are angles in the first quardrant such that $\tan \theta =\frac{1}{7}$ and $\sin \varphi =\frac{1}{\sqrt{10}}$ then $\theta +2\varphi ={45}^0$
Which of the above statements is correct?

• A. Only I
• B. Only II
• C. Both I&II
• D. Neither I nor II

Answer 1

The correct option is B Only II

Statement (I):

$\sin \alpha =\frac{12}{13}\forall \alpha \in (0,\frac{\pi }{2})⟹\cos \alpha =\sqrt{1-\frac{144}{169}}=\frac{5}{13}$

$\cos \beta =\frac{-3}{5}\forall \beta \in (\pi ,\frac{3\pi }{2})⟹\sin \alpha =-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$

$\therefore \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta =-\frac{56}{65}$

Statement (II):

$\sin \varphi =\frac{1}{\sqrt{10}}⟹\cos \varphi =\sqrt{1-\frac{1}{10}}=\frac{3}{\sqrt{10}}$

$\therefore \sin \left(2\varphi \right)=2\sin \varphi \cos \varphi =\frac{3}{5}$ and $\cos \left(2\varphi \right)={\cos }^2\varphi -{\sin }^2\varphi =\frac{4}{5}$

Also,

$\sin \theta =\frac{1}{\sqrt{1^2+7^2}}=\frac{1}{\sqrt{50}}$ and $\cos \theta =\frac{7}{\sqrt{50}}$

$\therefore \sin (\theta +2\varphi )=\sin \theta \cos \left(2\varphi \right)+\cos \theta \sin \left(2\varphi \right)=\frac{1}{\sqrt{2}}$.

Hence, only statement (II) is correct.