Solve the Following Question.(5 Marks)

Question 15 Marks

If $\cot \theta=\frac{3}{4}$ and $\pi<\theta<\frac{3 \pi}{2}$, then find the value of $4 \operatorname{cosec} \theta+5 \cos \theta$.

Answer

We know that,
$ \operatorname{cosec}^2 \theta=1+\cot ^2 \theta=\left(\frac{3}{4}\right)^2=1+\frac{9}{16}$
$\therefore \operatorname{cosec}^2 \theta=\frac{25}{16}$
$\therefore \operatorname{cosec} \theta= \pm \frac{5}{4}$
$\text { Since } \pi<\theta<\frac{3 \pi}{2} $
$\theta$ lies in the third quadrant.
$ \therefore \operatorname{cosec} \theta<0$
$\therefore \operatorname{cosec} \theta=-\frac{5}{4}$
$\cot \theta=\frac{3}{4}$
$\tan \theta=\frac{1}{\cot \theta}=\frac{4}{3} $
We know that,
$ \sec ^2 \theta=1+\tan ^2 \theta=1+\left(\frac{4}{3}\right)^2$
$=1+\frac{16}{9}=\frac{25}{9}$
$\therefore \sec \theta= \pm \frac{5}{3} $
Since $\theta$ lies in the third quadrant,
$ \sec \theta<0$
$\therefore \sec \theta=-\frac{5}{3}$
$\cos \theta=\frac{1}{\sec \theta}=\frac{-3}{5}$
$\therefore 4 \operatorname{cosec} \theta+5 \cos \theta$
$=4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)$
$=-5-3=-8 $
[Note: The question has been modified.]

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Question 25 Marks

If cosec θ + cot θ = 5, then evaluate sec θ.

Answer

$ \operatorname{cosec} \theta+\cot \theta=5$
$\therefore \frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5$
$\therefore \frac{1+\cos \theta}{\sin \theta}=5$
$\therefore 1+\cos \theta=5 \cdot \sin \theta $
Squaring both the sides, we get
$ 1+2 \cos \theta+\cos ^2 \theta=25 \sin ^2 \theta$
$\therefore \cos ^2 \theta+2 \cos \theta+1=25\left(1-\cos ^2 \theta\right)$
$\therefore \cos ^2 \theta+2 \cos \theta+1=25-25 \cos ^2 \theta$
$\therefore 26 \cos ^2 \theta+2 \cos \theta-24=0$
$\therefore 26 \cos ^2 \theta+26 \cos \theta-24 \cos \theta-24=0$
$\therefore 26 \cos \theta(\cos \theta+1)-24(\cos \theta+1)=0$
$\therefore(\cos \theta+1)(26 \cos \theta-24)=0$
$\therefore \cos \theta+1=\theta \text { or } 26 \cos \theta-24=0$
$\therefore \cos \theta=-1 \text { or } \cos \theta=\frac{24}{26}=\frac{12}{13}$
$\text { When } \cos \theta=-1, \sin \theta=0 $
$\therefore \cot \theta$ and $\operatorname{cosec} x$ are not defined,
$ \therefore \cos \theta \neq-1$
$\therefore \cos \theta=\frac{12}{13}$
$\therefore \sec \theta=\frac{1}{\cos \theta}=\frac{13}{12} $
[Note: Answer given in the textbook is -1 or $\frac{13}{12}$. However, as per our calculation it is only $\frac{13}{12}$.]

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Question 35 Marks

If $2 \sin A=1=\sqrt{2} \cos B$ and $\frac{\pi}{2}

Answer

Given, $2 \sin \mathrm{A}=1$
$\therefore \sin A=1 / 2$
we know that,
$ \cos ^2 \mathrm{~A}=1-\sin ^2 \mathrm{~A}=1-\left(\frac{1}{2}\right)^2=1-\frac{1}{4}=\frac{3}{4}$
$\therefore \cos \mathrm{A}= \pm \frac{\sqrt{3}}{2}$
$\text { Since } \frac{\pi}{2}<\mathrm{A}<\pi $
A lies in the 2 nd quadrant.
$\begin{aligned}
\therefore \quad \cos \mathrm{A} & = \pm \frac{\sqrt{3}}{2} \\
& \text { Since } \frac{\pi}{2}<\mathrm{A}<\pi,
\end{aligned}$
A lies in the $2^{\text {nd }}$ quadrant.
$\begin{aligned}
\therefore \quad \cos A & =-\frac{\sqrt{3}}{2} \\
\tan A & =\frac{\sin A}{\cos A}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}
\end{aligned}$
Also, $\sqrt{2} \cos \mathrm{B}=1$
$\therefore \quad \cos \mathrm{B}=\frac{1}{\sqrt{2}}$
We know that,
$\sin ^2 B=1-\cos ^2 B=1-\left(\frac{1}{\sqrt{2}}\right)^2 \frac{1}{2}=\frac{1}{2}$
$\therefore \sin B= \pm \frac{1}{\sqrt{2}}$
Since $\frac{3 \pi}{2}$B lies in the 4th quadrant,
$ \therefore \quad \sin \mathrm{B}=-\frac{1}{\sqrt{2}}$
$\tan B=\frac{\sin B}{\cos B}=\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1$
$\therefore \quad \frac{\tan A+\tan B}{\cos A-\cos B}=\frac{-\frac{1}{\sqrt{3}}-1}{-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}=\frac{\frac{-1-\sqrt{3}}{\sqrt{3}}}{\frac{-\sqrt{3}-\sqrt{2}}{2}}$
$=\frac{-(1+\sqrt{3})}{\sqrt{3}} \times \frac{2}{-(\sqrt{3}+\sqrt{2})}$
$=\frac{2(1+\sqrt{3})}{\sqrt{3}(\sqrt{3}+\sqrt{2})}$

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Maths Solve the Following Question.(5 Marks)

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