Question

$sin\theta =\frac{12}{13},cos\varnothing =\frac{8}{17}$ find $tan2\theta ,cot\frac{\varnothing }{2},sec(\theta +\varphi )$

Answer 1

$\sin \theta =\frac{12}{13}.....\left(1\right)\left(\text{Given}\right)$

As we know that,

$\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}.....\left(2\right)$

Compring equation $\left(1\right)\&\left(2\right)$, we get

Perpendicular $\left(p\right)=12$

Hypotenuse $\left(h\right)=13$

Base $\left(b\right)=?$

By pythagoras theorem, we have

$h^2=p^2+b^2$

${\left(13\right)}^2={\left(12\right)}^2=b^2$

$\Rightarrow b=\sqrt{169-144}=5$

Therefore,

$\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

$\Rightarrow \tan \theta =\frac{12}{5}$

Now, as we know that,

$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\therefore \tan 2\theta =\frac{2\left(\frac{12}{5}\right)}{1-{\left(\frac{12}{5}\right)}^2}$

$\Rightarrow \tan 2\theta =\frac{\left(\frac{24}{5}\right)}{\left(\frac{25-144}{25}\right)}$

$\Rightarrow \tan 2\theta =-\frac{120}{119}$

Again as we know that,

$\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}$

$\Rightarrow \cos \theta =\frac{5}{13}$

$\cos \varphi =\frac{8}{17}.....\left(3\right)\left(\text{Given}\right)$

As we know that,

$\cos \varphi =\frac{\text{Base}}{\text{Hypotenuse}}.....\left(4\right)$

Compring equation $\left(3\right)\&\left(4\right)$, we get

Perpendicular $\left(p\right)=?$

Hypotenuse $\left(h\right)=17$

Base $\left(b\right)=8$

By pythagoras theorem, we have

$h^2=p^2+b^2$

${\left(17\right)}^2=p^2+{\left(8\right)}^2$

$\Rightarrow b=\sqrt{289-64}=15$

Therefore,

$\sin \varphi =\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{15}{17}$

Now, as we know that,

$\cot \left(\frac{\varphi }{2}\right)=\frac{1+\cos \varphi }{\sin \varphi }$

$\therefore \cot \left(\frac{\varphi }{2}\right)=\frac{1+\frac{8}{17}}{\frac{15}{17}}$

$\Rightarrow \cot \left(\frac{\varphi }{2}\right)=\frac{17+8}{15}=\frac{25}{15}=\frac{5}{3}$

Now,

$\sec (\theta +\varphi )$

$=\frac{1}{\cos (\theta +\varphi )}$

$=\frac{1}{\cos \varphi \cos \theta -\sin \theta \sin \varphi }[\because \cos (A+B)=\cos A\cos B-\sin A\sin B]$

$=\frac{1}{\left(\frac{5}{13}\right)\left(\frac{8}{17}\right)-\left(\frac{12}{13}\right)\left(\frac{15}{17}\right)}$

$=\frac{13\times 17}{40-180}$

$=-\frac{221}{140}$

Hence, the answer is-

$\tan 2\theta =-\frac{120}{119}$

$\cot \left(\frac{\varphi }{2}\right)=\frac{5}{3}$

$sec(\theta +\varphi )=-\frac{221}{140}$