Solve the Following Question.(4 Marks)

Question 14 Marks

$\frac{\tan ^2 x}{1+\tan ^2 x}+\frac{\cot ^3 x}{1+\cot ^2 x}=\sec x \operatorname{cosec} x-2 \sin x \cos x$

Answer

$\begin{aligned}
& \text { L.H.S. }=\frac{\tan ^3 x}{1+\tan ^2 x}+\frac{\cot ^3 x}{1+\cot ^2 x} \\
& =\frac{\left(\frac{\sin ^3 x}{\cos ^3 x}\right)}{\sec ^2 x}+\frac{\frac{\cos ^3 x}{\sin ^3 x}}{\operatorname{cosec}^2 x} \\
& =\frac{\sin ^3 x}{\cos ^3 x} \times \cos ^2 x+\frac{\cos ^3 x}{\sin ^3 x} \times \sin ^2 x \\
& =\frac{\sin ^3 x}{\cos x}+\frac{\cos ^3 x}{\sin x} \\
& =\frac{\sin ^4 x+\cos ^4 x}{\cos x \sin x} \\
& =\frac{\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^2}{\cos x \sin x} \\
& =\frac{\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x}{\cos x \sin x} \\
& \cdots\left[\because a^2+b^2=(a+b)^2-2 a b\right] \\
& =\frac{1^2-2 \sin ^2 x \cos ^2 x}{\cos x \sin x} \\
& =\frac{1}{\cos x \sin x}-\frac{2 \sin ^2 x \cos ^2 x}{\cos x \sin x} \\
& =\sec x \cos x-2 \sin x \cos x \\
& =\mathrm{R} . \mathrm{H} \cdot \mathrm{S} \text {. } \\
\end{aligned}$

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Question 24 Marks

In any triangle $\mathrm{ABC}_r \sin \mathrm{A}-\cos \mathrm{B}=\cos \mathrm{C}$. Show that $\angle \mathrm{B}=\frac{\pi}{2}$.

Answer

$ \sin A-\cos B=\cos C$
$ \therefore \sin A=\cos B+\cos C$
$ \therefore 2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}=2 \cos \left(\frac{\mathrm{~B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$ \therefore 2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}=2 \cos \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\left[\begin{array}{l}\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi, \\ \therefore \frac{\mathrm{B}+\mathrm{C}}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2}\end{array}\right]$
$\therefore 2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}=2 \sin \frac{\mathrm{~A}}{2} \cos \left(\frac{\mathrm{~B}-\mathrm{C}}{2}\right)$
$\therefore \cos \frac{\mathrm{A}}{2}=\cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right) \ldots\left[\because \sin \frac{\mathrm{A}}{2} \neq 0\right]$
$ \therefore \frac{\mathrm{A}}{2}=\frac{\mathrm{B}-\mathrm{C}}{2}$
$ A=B-C$
$ \text { In } \triangle \mathrm{ABC}$
$ A+B+C=\pi$
$ \therefore B-C+B+C=\pi$
$ \therefore 2 B=\pi$
$ \therefore B=\frac{\pi}{2}$

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Question 34 Marks

If $A+B+C=\frac{3 \pi}{2}$, then $\cos 2 A+\cos 2 B+\cos 2 C=1-4 \sin A$ $\sin B \sin C$

Answer

$\begin{aligned}
& \text { In } \Delta \mathrm{ABC}, \\
& \mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{3 \pi}{2} \\
& \therefore \quad \mathrm{A}+\mathrm{B}=\frac{3 \pi}{2}-\mathrm{C} \\
& \therefore \quad \cos (\mathrm{A}+\mathrm{B})=\cos \left(\frac{3 \pi}{2}-\mathrm{C}\right) \\
&=-\sin \mathrm{C} \\
& \text { L.H.S. }=\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C} \\
&= 2 \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)+\cos 2 \mathrm{C} \\
&= 2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+\cos 2 \mathrm{C} \\
&= 2(-\sin \mathrm{C}) \cos (\mathrm{A}-\mathrm{B})+1-2 \sin { }^2 \mathrm{C} \\
&= 1-2 \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})+\sin \mathrm{C}] \\
&= 1-2 \sin \mathrm{C}\{\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})\} \\
&= 1-2 \sin \mathrm{C}\\
& \times 2 \sin \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{A}+\mathrm{B}-\mathrm{A}+\mathrm{B}}{2}\right) \\
&= 1-2 \sin \mathrm{C}(2 \sin \mathrm{A} \sin \mathrm{B}) \\
&= 1-4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C} \\
&= \mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S} .
\end{aligned}$

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Question 44 Marks

$\tan \mathrm{A}+\tan \left(60^{\circ}+\mathrm{A}\right)+\tan \left(120^{\circ}+\mathrm{A}\right)=3 \tan 3 \mathrm{~A}$

Answer

L.H.S.
$\begin{aligned}
& =\tan \mathrm{A}+\tan \left(60^{\circ}+\mathrm{A}\right)+\tan \left(120^{\circ}+\mathrm{A}\right) \\
& =\tan A+\frac{\tan 60^{\circ}+\tan A}{1-\tan 60^{\circ} \tan A}+\frac{\tan 120^{\circ}+\tan A}{1-\tan 120^{\circ} \tan A} \\
& =\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3} \tan A}+\frac{-\tan 60^{\circ}+\tan A}{1-\left(-\tan 60^{\circ}\right) \tan A} \\
& \cdots\left[\begin{array}{rl}
\because \tan 120^{\circ} & =\tan \left(180^{\circ}-60^{\circ}\right) \\
& =-\tan 60^{\circ}
\end{array}\right] \\
& =\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3} \tan A}-\frac{\sqrt{3}-\tan A}{1+\sqrt{3} \tan A} \\
& =\tan \mathrm{A}\\
& +\frac{\sqrt{3}+3 \tan A+\tan A+\sqrt{3} \tan ^2 A-\sqrt{3}+\tan A+3 \tan A-\sqrt{3} \tan ^2 A}{(1-\sqrt{3} \tan A)(1+\sqrt{3} \tan A)} \\
& =\tan A+\frac{8 \tan A}{1-3 \tan ^2 A} \\
& =\frac{\tan A-3 \tan ^3 \mathrm{~A}+8 \tan \mathrm{A}}{1-3 \tan ^2 \mathrm{~A}} \\
& =\frac{9 \tan A-3 \tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}} \\
& =3\left(\frac{3 \tan A-\tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}}\right) \\
& =3 \tan 3 \mathrm{~A} \\
& =\text { R.H.S. } \\
\end{aligned}$

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Question 54 Marks

$\frac{2 \cos 2 \mathrm{~A}+1}{2 \cos 2 \mathrm{~A}-1}=\tan \left(60^{\circ}+\mathrm{A}\right) \tan \left(60^{\circ}-\mathrm{A}\right) $

Answer

\begin{aligned}
& \text { R.H.S. }=\tan \left(60^{\circ}+\mathrm{A}\right) \tan \left(60^{\circ}-\mathrm{A}\right) \\
& =\frac{\sin \left(60^{\circ}+\mathrm{A}\right) \sin \left(60^{\circ}-\mathrm{A}\right)}{\cos \left(60^{\circ}+\mathrm{A}\right) \cos \left(60^{\circ}-\mathrm{A}\right)} \\
& =\frac{2 \sin \left(60^{\circ}+\mathrm{A}\right) \sin \left(60^{\circ}-\mathrm{A}\right)}{2 \cos \left(60^{\circ}+\mathrm{A}\right) \cos \left(60^{\circ}-\mathrm{A}\right)} \\
& =\frac{\cos \left[60^{\circ}+\mathrm{A}-\left(60^{\circ}-\mathrm{A}\right)\right]-\cos \left(60^{\circ}+\mathrm{A}+60^{\circ}-\mathrm{A}\right)}{\cos \left(60^{\circ}+\mathrm{A}+60^{\circ}-\mathrm{A}\right)+\cos \left[60^{\circ}+\mathrm{A}-\left(60^{\circ}-\mathrm{A}\right)\right]} \\
& =\frac{\cos 2 \mathrm{~A}-\cos 120^{\circ}}{\cos 120^{\circ}-\cos 2 \mathrm{~A}} \\
& =\frac{\cos 2 \mathrm{~A}-\cos \left(180^{\circ}-60^{\circ}\right)}{\cos \left(180^{\circ}-60^{\circ}\right)+\cos 2 \mathrm{~A}} \\
& =\frac{\cos 2 \mathrm{~A}-\left(-\cos 60^{\circ}\right)}{-\cos 60^{\circ}+\cos 2 \mathrm{~A}} \\
& =\frac{\cos 2 \mathrm{~A}+\frac{1}{2}}{-\frac{1}{2}+\cos 2 \mathrm{~A}} \\
& =\frac{2 \cos 2 \mathrm{~A}+1}{2 \cos 2 \mathrm{~A}-1} \\
& =\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{S} .
\end{aligned}

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Question 64 Marks

$\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}=-\frac{1}{2}$

Answer

L.H.S.
$\begin{aligned}
= & \cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} \\
= & \left(\cos 132^{\circ}+\cos 12^{\circ}\right)+\left(\cos 156^{\circ}+\cos 84^{\circ}\right) \\
= & 2 \cos \left(\frac{132^{\circ}+12^{\circ}}{2}\right) \cos \left(\frac{132^{\circ}-12^{\circ}}{2}\right) \\
& +2 \cos \left(\frac{156^{\circ}+84^{\circ}}{2}\right) \cos \left(\frac{156^{\circ}-84^{\circ}}{2}\right) \\
= & 2 \cos 72^{\circ} \cos 60^{\circ}+2 \cos 120^{\circ} \cos 36^{\circ} \\
= & 2 \cos 72^{\circ} \cos 60^{\circ}+2 \cos \left(180^{\circ}-60^{\circ}\right) \cos 36^{\circ} \\
= & 2 \cos 72^{\circ} \cos 60^{\circ}+2\left(-\cos 60^{\circ}\right) \cos 36^{\circ} \\
= &2 \cos 72^{\circ}\left(\frac{1}{2}\right)-2\left(\frac{1}{2}\right) \cos 36^{\circ} \\
= & \cos 72^{\circ}-\cos 36^{\circ} \\
= & 2 \sin \left(\frac{72^{\circ}+36^{\circ}}{2}\right) \sin \left(\frac{36^{\circ}-72^{\circ}}{2}\right) \\
= & 2 \sin 54^{\circ} \sin \left(-18^{\circ}\right) \\
= & -2 \sin 54^{\circ} \cdot \sin 18^{\circ} \\
= & -2\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right) \\
= & -\frac{1}{8}(5-1)\\
= & -\frac{1}{2}=\text { R.H.S. }
\end{aligned}$

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Question 74 Marks

$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}$

Answer

L.H.S.
$=\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$
Since, $\cos (\pi-\theta)=-\cos \theta$
$\begin{aligned}
& \therefore \quad \cos \frac{7 \pi}{8}=\cos \left(\pi-\frac{\pi}{8}\right)=-\cos \frac{\pi}{8} \\
& \quad \text { and } \cos \frac{5 \pi}{8}=\cos \left(\pi-\frac{3 \pi}{8}\right)=-\cos \frac{3 \pi}{8} \\
& \therefore \quad \text { L.H.S. }=\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)
\end{aligned}$
$\left(1-\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)$
...[From (i) and (ii)]
$\begin{aligned}
& =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right) \\
& =\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8} \\
& =\frac{1}{4}\left(2 \sin \frac{\pi}{8} \sin \frac{3 \pi}{8}\right)^2 \\
& =\frac{1}{4}\left[\cos \left(\frac{\pi}{8}-\frac{3 \pi}{8}\right)-\cos \left(\frac{\pi}{8}+\frac{3 \pi}{8}\right)\right]^2 \\
& =\frac{1}{4}\left[\cos \left(-\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{2}\right)\right]^2 \\
& =\frac{1}{4}\left(\cos \left(\frac{\pi}{4}\right)-0\right)^2=\frac{1}{4}\left(\frac{1}{\sqrt{2}}\right)^2 \\
& =\frac{1}{4}\left(\frac{1}{2}\right) \\
& =\frac{1}{8} \\
& =\text { R. H.S. } \\
\end{aligned}$

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Question 84 Marks

Find $\sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}$ if $\tan x=\frac{4}{3}, x$ lies in II quadrant.

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Question 94 Marks

If $\sin A+\sin B=x$ and $\cos A+\cos B=y$ then show that $\sin (A+B)=\frac{2 x y}{x^2+y^2}$

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Question 104 Marks

$\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}=2 \sin \mathrm{A} \sin \mathrm{B} \cos \mathrm{C}$

Answer

We know that, $\sin ^2=\frac{1-\cos 2 \theta}{2}$
L.H.S.
$
\begin{aligned}
& =\sin ^2+\sin ^2 B+\sin ^2 C \\
& =\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2}-\sin ^2 C \\
& =\frac{1}{2}[2-(\cos 2 A+\cos 2 B)]-\sin ^2 C \\
& =\frac{1}{2}\left[2-2 \cdot \cos \left(\frac{2 A+2 B}{2}\right) \cdot \cos \left(\frac{2 A-2 B}{2}\right)\right]
\end{aligned}$
$\begin{aligned}
& -\sin ^2 C \\
& =1-\cos (A+B) \cdot \cos (A-B)-\sin ^2 C \\
& =\left(1-\sin ^2 C\right)-\cos (A+B) \cdot \cos (A-B) \\
& =\cos ^2 C-\cos (A+B) \cdot \cos (A-B) \\
& \therefore \cos (A+B)=\cos (\text { it }-C) \\
& \therefore \cos (A+B)=-\cos C \ldots \text { (i) } \\
& \therefore \text { L.H.S. }=\cos ^2 \mathrm{C}+\cos \mathrm{C} \cdot \cos (\mathrm{A}-\mathrm{B}) \\
& \text {... [From (i)] } \\
& =\cos C[\cos C+\cos (A-B)] \\
& =\cos C[-\cos (A+B)+\cos (A-B)] \\
& \text {... [From (i)] } \\
& =\cos C[\cos (A-B)-\cos (A+B)] \\
& =\cos C(2 \sin A \cdot \sin B) \\
& =2 \sin A \cdot \sin B \cdot \cos C \\
& =\mathrm{R} . \mathrm{H} . \mathrm{S} \text {. } \\
\end{aligned}$
[Note: The question has been modified.]

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Question 114 Marks

In $\triangle A B C, A+B+C=\pi$, show that
$\cos 2 A+\cos 2 B+\cos 2 C=-1-4 \cos A \cos B \cos C$

Answer

$\begin{aligned}
& \text { L.H.S. }=\cos 2 A+\cos 2 B+\cos 2 C \\
& =2 \cdot \cos \left(\frac{2 A+2 B}{2}\right) \cdot \cos \left(\frac{2 A-2 B}{2}\right)+\cos 2 C \\
& =2 \cdot \cos (A+B) \cdot \cos (A-B)+2 \cos ^2 C-1 \\
& \text { In } \triangle A B C, A+B+C=\pi \\
& \therefore A+B=\pi-C \\
& \therefore \cos (A+B)=\cos (\pi-C) \\
& \therefore \cos (A+B)=-\cos C \ldots \ldots \ldots \text { (i) }
\end{aligned}$
$\begin{aligned}
& \therefore \text { L.H.S. }=-2 \cdot \cos C \cdot \cos (A-B)+2 \cdot \cos ^2 C-1 \ldots[\text { From }(i)] \\
& =-1-2 \cdot \cos C \cdot[\cos (A-B)-\cos C] \\
& =-1-2 \cdot \cos C \cdot[\cos (A-B)+\cos (A+B)] \\
& \ldots[\text { From (i)] } \\
& =-1-2 \cdot \cos C \cdot(2 \cdot \cos A \cdot \cos B) \\
& =-1-4 \cdot \cos A \cdot \cos B \cdot \cos C=\text { R.H.S. }
\end{aligned}$

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Question 124 Marks

Prove the following : $\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=3 / 16$

Answer

$\begin{aligned}
& \text { L.H.S. }=\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 60^{\circ} \cdot \sin 80^{\circ} \\
& =\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot\left(\frac{\sqrt{3}}{2}\right) \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\left(2 \cdot \sin 40^{\circ} \cdot \sin 20^{\circ}\right) \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{4}\left[\cos \left(40^{\circ}-20^{\circ}\right)-\cos \left(40^{\circ}+20^{\circ}\right)\right] \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{4}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{4} \cdot \cos 20^{\circ} \cdot \sin 80^{\circ}-\frac{\sqrt{3}}{4} \cos 60^{\circ} \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{2 \times 4}\left(2 \sin 80^{\circ} \cdot \cos 20^{\circ}\right)-\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right) \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \cdot\left[\sin \left(80^{\circ}+20^{\circ}\right)+\sin \left(80^{\circ}-20^{\circ}\right)\right]
\end{aligned}$
$-\frac{\sqrt{3}}{8} \cdot \sin 80^{\circ}$
$\begin{aligned}
& =\frac{\sqrt{3}}{8} \cdot\left(\sin 100^{\circ}+\sin 60^{\circ}\right)-\frac{\sqrt{3}}{8} \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \sin 100^{\circ}+\frac{\sqrt{3}}{8} \sin 60^{\circ}-\frac{\sqrt{3}}{8} \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \cdot \sin \left(180^{\circ}-80^{\circ}\right)+\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{8} \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \sin 80^{\circ}+\frac{3}{16}-\frac{\sqrt{3}}{8} \sin 80^{\circ} \\
& =\frac{3}{16}=\text { R.H.S. } \quad \ldots\left[\because \sin \left(180^{\circ}-\theta\right)=\sin \theta\right]
\end{aligned}$

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Question 134 Marks

Prove the following : $\sec 840^{\circ} \cot \left(-945^{\circ}\right)+\sin 600^{\circ} \tan \left(-690^{\circ}\right)=3 / 2$

Answer

\begin{aligned}
& \sec 840^{\circ}=\sec \left(720^{\circ}+120^{\circ}\right) \\
& =\sec \left(2 \times 360^{\circ}+120^{\circ}\right) \\
& =\sec \left(120^{\circ}\right) \\
& =\sec \left(90^{\circ}+30^{\circ}\right) \\
& =-\operatorname{cosec} 30^{\circ} \\
& =-2 \\
& \cot \left(-945^{\circ}\right)=-\cot 945^{\circ} \\
& =-\cot \left(720^{\circ}+225^{\circ}\right) \\
& =-\cot \left(2 \times 360^{\circ}+225^{\circ}\right) \\
& =-\cot \left(225^{\circ}\right) \\
& =-\cot \left(180^{\circ}+459\right) \\
& =-\cot 45^{\circ} \\
& =-1 \\
& \sin 600^{\circ}=\sin \left(360^{\circ}+240^{\circ}\right) \\
& =\sin \left(240^{\circ}\right) \\
& =\sin \left(180^{\circ}+60^{\circ}\right) \\
& =-\sin 60^{\circ}=-\frac{\sqrt{3}}{2} \\
& \tan \left(-690^{\circ}\right)=-\tan 690^{\circ} \\
& =-\tan \left(360^{\circ}+330^{\circ}\right) \\
& =-\tan \left(330^{\circ}\right) \\
& =-\tan \left(360^{\circ}-30^{\circ}\right) \\
& =-\left(-\tan 30^{\circ}\right) \\
& =\tan 30^{\circ} 0=\frac{1}{\sqrt{3}} \\
& \text { L.H.S. }=\sec 840^{\circ} \cot \left(-945^{\circ}\right)+\sin 600^{\circ} \tan \left(-690^{\circ}\right) \\
& =(-2)(-1)+\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{3}}\right) \\
& =2-\frac{1}{2}=\frac{3}{2} \\
& =\mathrm{R}, \mathrm{H} . \mathrm{S} \text {. } \\
&
\end{aligned}

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Maths Solve the Following Question.(4 Marks)

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