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Maths Solve the Following Question.(5 Marks)

Trigonometry – II · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 10 questions.

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Solve the Following Question.(5 Marks)

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10 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
In $\triangle \mathrm{ABC}, \angle \mathrm{C}=\frac{2 \pi}{3}$, then prove that $\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}-\cos \mathrm{A} \cos \mathrm{B}=\frac{3}{4}$.
Answer
$\begin{aligned}
& \angle C=\frac{2 \pi}{3} \\
& \text { In } \triangle \mathrm{ABC} \text {, } \\
& \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\
& \therefore \quad \mathrm{A}+\mathrm{B}+\frac{2 \pi}{3}=\pi \\
& \therefore \quad \mathrm{A}+\mathrm{B}=\frac{\pi}{3} \\
& \therefore \quad \cos (A+B)=\cos \frac{\pi}{3}=\frac{1}{2} \\
& \text { L.H.S. }=\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}-\cos \mathrm{A} \cos \mathrm{B} \\
& =\frac{1}{2}\left(2 \cos ^2 A+2 \cos ^2 B\right)-\frac{1}{2}(2 \cos A \cos B) \\
& =\frac{1}{2}(1+\cos 2 \mathrm{~A}+1+\cos 2 \mathrm{~B}) \\
& -\frac{1}{2}(2 \cos A \cos B) \\
& =\frac{1}{2}\left[2+2 \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\right] \\
& -\frac{1}{2}[\cos (A+B)+\cos (A-B)] \\
& =\frac{1}{2}[2+2 \cos (A+B) \cos (A-B)] \\
& -\frac{1}{2}[\cos (A+B)+\cos (A-B)] \\
& =\frac{1}{2}\left[2+2\left(\frac{1}{2}\right) \cos (A-B)\right] \\
& -\frac{1}{2}\left[\frac{1}{2}+\cos (A-B)\right] \\
& \text {...[From (i)] } \\
& =\frac{1}{2}\left[2+\cos (A-B)-\frac{1}{2}-\cos (A-B)\right] \\
& =\frac{1}{2}\left(\frac{3}{2}\right)=\frac{3}{4}\\
& =\text { R.H.S. } \\
\end{aligned}$
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Question 25 Marks
$\tan A+2 \tan 2 A+4 \tan 4 A+8 \cot 8 \mathrm{~A}=\cot A $
Answer
\begin{aligned}
& \text { We have to prove that, } \\
& \tan A+2 \tan 2 A+4 \tan 4 A+8 \cot 8 A=\cot A \\
& \text { i.e., to prove, } \\
& \cot A-\tan A-2 \tan 2 A-4 \tan 4 A-8 \cot 8 A=0 \\
& \cot \theta-\tan \theta=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta} \\
& =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos \theta \sin \theta} \\
& =\frac{2 \cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos 2 \theta}{\sin 2 \theta} \\
& \therefore \cot \theta-\tan \theta=2 \cot 2 \theta \ldots \text {...(i) } \\
& \text { L.H.S. }=\cot A-\tan A-2 \tan 2 A-4 \tan 4 A-8 \cot 8 A \\
& =2 \cot 2 A-2 \tan 2 A-4 \tan 4 A-8 \cot 8 A \ldots . .[\text { From (i)] } \\
& =2(\cot 2 A-\tan 2 A)-4 \tan 4 A-8 \cot 8 A \\
& =2 \times 2 \cot 2(2 A)-4 \tan 4 A-8 \cot 8 A \ldots \ldots[\text { [rom (i)] } \\
& =4(\cot 4 \mathrm{~A}-\tan 4 \mathrm{~A})-8 \cot 8 \mathrm{~A} \\
& =4 \times 2 \cot 2(4 \mathrm{~A})-8 \cot 8 \mathrm{~A} \ldots \ldots . \text { [From (i)] } \\
& =8 \cot 8 \mathrm{~A}-8 \cot 8 \mathrm{~A}=0 \\
& =\text { R.H.S. } \\
\end{aligned}
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Question 35 Marks
$3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4\left(\sin ^6 x+\cos ^6 x\right)=13 $
Answer
\begin{aligned}
& (\sin x-\cos x)^4 \\
& =\left[(\sin x-\cos x)^2\right]^2 \\
& =\left(\sin ^2 x+\cos ^2 x-2 \sin x \cos x\right)^2 \\
& =(1-2 \sin x \cos x)^2 \\
& =1-4 \sin x \cos x+4 \sin ^2 x \cos ^2 x \\
& (\sin x+\cos x)^2=\sin ^2 x+\cos ^2 x+2 \sin x \cos x=1+2 \sin x \cos x \\
& \sin ^6 x+\cos ^6 x \\
& =\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3 \\
& =\left(\sin { }^2 x+\cos ^2 x\right)^3-3 \sin ^2 x \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right) \ldots . . \because a^3+ \\
& \left.b^3=(a+b)^3-3 a b(a+b)\right] \\
& =1^3-3 \sin ^2 x \cos ^2 x(1) \\
& =1-3 \sin ^2 x \cos ^2 x \\
& \text { L.H.S. }=3\left(\sin ^2 x-\cos x\right)^4+6(\sin x+\cos x)^2+4\left(\sin { }^6 x+\cos ^6 x\right) \\
& =3\left(1-4 \sin ^2 x \cos ^2 x+4 \sin ^2 x \cos ^2 x\right)+6\left(1+2 \sin ^2 x \cos x\right)+ \\
& 4\left(1-3 \sin ^2 x \cos ^2 x\right) \\
& =3-12 \sin ^2 x \cos ^2 x+12 \sin ^2 x \cos ^2 x+6+12 \sin ^2 x \cos x+4- \\
& 12 \sin ^2 x \cos ^2 x \\
& =13 \\
& =\text { R.H.S. }
\end{aligned}
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Question 45 Marks
$ \text { cosec } 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ}=0 $
Answer
$\begin{aligned} & \text { L.H.S. }=\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ} \\ & =\operatorname{cosec} 48^{\circ}+\operatorname{cosec}\left(180^{\circ}-84^{\circ}\right)+\operatorname{cosec}\left(180^{\circ}+12^{\circ}\right)+\operatorname{cosec}\left(360^{\circ}+24^{\circ}\right) \\ & =\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 84^{\circ}+\operatorname{cosec}\left(-12^{\circ}\right)+\operatorname{cosec} 24^{\circ} \\ & =\frac{1}{\sin 48^{\circ}}+\frac{1}{\sin 84^{\circ}}+\frac{1}{-\sin 12^{\circ}}+\frac{1}{\sin 24^{\circ}} \\ & =\left(\frac{1}{\sin 48^{\circ}}-\frac{1}{\sin 12^{\circ}}\right)+\left(\frac{1}{\sin 84^{\circ}}+\frac{1}{\sin 24^{\circ}}\right) \\ & =-\frac{\left(\sin 48^{\circ}-\sin 12^{\circ}\right)}{\sin 48^{\circ} \sin 12^{\circ}}+\frac{\left(\sin 84^{\circ}+\sin 24^{\circ}\right)}{\sin 84^{\circ} \sin 24^{\circ}} \\ & =-\frac{2 \cos \left(\frac{48^{\circ}+12^{\circ}}{2}\right) \sin \left(\frac{48^{\circ}-12^{\circ}}{2}\right)}{\frac{1}{2}\left[\cos \left(48^{\circ}-12^{\circ}\right)-\cos \left(48^{\circ}+12^{\circ}\right)\right]} \\ & +\frac{2 \sin \left(\frac{84^{\circ}+24^{\circ}}{2}\right) \cos \left(\frac{84^{\circ}-24^{\circ}}{2}\right)}{\frac{1}{2}\left[\cos \left(84^{\circ}-24^{\circ}\right)-\cos \left(84^{\circ}+24^{\circ}\right)\right]} \\ & =-\frac{2 \cos 30^{\circ} \sin 18^{\circ}}{\frac{1}{2}\left(\cos 36^{\circ}-\cos 60^{\circ}\right)} \\ & +\frac{2 \sin 54^{\circ} \cos 30^{\circ}}{\frac{1}{2}\left(\cos 60^{\circ}-\cos 108^{\circ}\right)} \\ & =\frac{4 \cos 30^{\circ} \sin 18^{\circ}}{\cos 60^{\circ}-\cos 36^{\circ}}+\frac{4 \sin 54^{\circ} \cos 30^{\circ}}{\cos 60^{\circ}+\sin 18^{\circ}} \\ & =4 \cos 30^{\circ}\left[\frac{\sin 18^{\circ}}{\cos 60^{\circ}-\cos 36^{\circ}}+\frac{\sin 54^{\circ}}{\cos 60^{\circ}+\sin 18^{\circ}}\right] \\ & =4 \cos 30^{\circ}\left[\frac{\sin 18^{\circ}}{\cos 60^{\circ}-\cos 36^{\circ}}+\frac{\cos 36^{\circ}}{\cos 60^{\circ}+\sin 18^{\circ}}\right] \\ & =4 \cos 30^{\circ}\left[\frac{\frac{\sqrt{5}-1}{4}}{\frac{1}{2}-\frac{\sqrt{5}+1}{4}}+\frac{\frac{\sqrt{5}+1}{4}}{\frac{1}{2}+\frac{\sqrt{5}-1}{4}}\right] \\ & =4 \cos 30^{\circ}(-1+1)=0=\text { R.H.S. } \\ & \end{aligned}$
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Question 55 Marks
$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}$
Answer
$\begin{aligned}
& \text { L.H.S. }=\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15} \\
& =\frac{1}{2 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{2 \pi}{15} \cos \frac{2 \pi}{15}\right)
\end{aligned}$
$\begin{aligned}
& =\frac{1}{2 \sin \frac{2 \pi}{15}}\left(\sin \frac{4 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}\right) \\
\end{aligned}$
$\ldots[\because 2 \sin \theta \cos \theta=\sin 2 \theta]$
$=\frac{1}{2 \times 2 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{4 \pi}{15} \cos \frac{4 \pi}{15}\right) \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$
$=\frac{1}{4 \sin \frac{2 \pi}{15}}\left(\sin \frac{8 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}\right)$
$\begin{aligned}
& =\frac{1}{2 \times 4 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{8 \pi}{15} \cos \frac{8 \pi}{15}\right) \cos \frac{16 \pi}{15} \\
& =\frac{1}{8 \sin \frac{2 \pi}{15}}\left(\sin \frac{16 \pi}{15} \cos \frac{16 \pi}{15}\right) \\
& =\frac{1}{2 \times 8 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{16 \pi}{15} \cos \frac{16 \pi}{15}\right) \\
& =\frac{1}{16 \sin \frac{2 \pi}{15}} \sin \left(\frac{32 \pi}{15}\right) \\
& =\frac{1}{16 \sin \frac{2 \pi}{15}} \sin \left(2 \pi+\frac{2 \pi}{15}\right) \\
& =\frac{1}{16 \sin \frac{2 \pi}{15}}\left(\sin \frac{2 \pi}{16}\right) \ldots[\because \sin (2 \pi+\theta)=\sin \theta] \\
& =\frac{1}{16} \\
& =R.H.S
\end{aligned}$
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Question 65 Marks
In $\triangle A B C$ prove that
$\frac{\cos A-\cos B+\cos C+1}{\cos A+\cos B+\cos C-1}=\cot \frac{A}{2} \cot \frac{C}{2}$
Answer
$\begin{aligned} & \text { L.H.S. }=\frac{\cos A-\cos B+\cos C+1}{\cos A+\cos B+\cos C-1} \\ & =\frac{[\cos A-\cos B]+[1+\cos C]}{[\cos A+\cos B]-[1-\cos C]} \\ & =\frac{2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)+2 \cos ^2 \frac{C}{2}}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\left(-2 \sin ^2 \frac{C}{2}\right)} \\ & =\frac{2 \sin \left(\frac{\pi}{2}-\fraaac{c}{2}\right) \sin \left(\frac{B-A}{2}\right)+2 \cos ^2 \frac{C}{2}}{2 \cos \left(\frac{\pi}{2}-\frac{c}{2}\right) \cos \left(\frac{A-B}{2}\right)+\left(-2 \sin ^2 \frac{C}{2}\right)} \\ & =\frac{2 \cos \frac{C}{2} \sin \left(\frac{B-A}{2}\right)+2 \cos ^2 \frac{C}{2}}{2 \sin \frac{C}{2} \cos \left(\frac{A+B}{2}\right)-2 \sin ^2 \frac{C}{2}} \\ & =\frac{\cos \frac{C}{2}\left[\sin \left(\frac{B-A}{2}\right)+\cos \frac{C}{2}\right]}{\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right.} \\ & =\cot \frac{C}{2} \frac{\left[\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{B-A}{2}\right)\right]}{\left[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right]} \\ & =\cot \frac{\mathrm{C}}{2} \frac{\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{B-A}{2}\right)}{\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)} \\ & =\cot \frac{C}{2} \cdot \frac{2 \sin \frac{\left(\frac{A+B}{2}+\frac{B-A}{2}\right)}{2} \cos \frac{\left(\frac{A+B}{2}-\frac{B-A}{2}\right)}{2}}{2 \sin \frac{\left(\frac{A-B}{2}+\frac{A+B}{2}\right)}{2} \sin \frac{\left(\frac{A+B}{2}-\frac{A-B}{2}\right)}{2}} \\ & \end{aligned}$$\begin{aligned} & =\cot \frac{\mathrm{C}}{2} \cdot \frac{2 \sin \frac{\mathrm{B}}{2} \cos \frac{\mathrm{A}}{2}}{2 \sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2}} \\ & =\cot \frac{\mathrm{C}}{2} \frac{\cos \frac{\mathrm{A}}{2}}{\sin \frac{\mathrm{A}}{2}} \\ & =\cot \frac{\mathrm{C}}{2} \cot \frac{\mathrm{A}}{2} \\ & =\text { R.H.S. }\end{aligned}$
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Question 75 Marks
In $\triangle A B C_1 A+B+C=\pi$ show that $\cos ^2 A+\cos ^2 B-\cos ^2 C=1-2 \sin A \sin B \cos C$
Answer
we know that $\cos ^2 \theta=\frac{1+\cos 2 \theta}{2}$
L.H.S.
$\begin{aligned}
& =\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C} \\
& =\frac{1+\cos 2 \mathrm{~A}}{2}+\frac{1+\cos 2 \mathrm{~B}}{2}-\cos ^2 \mathrm{C} \\
& =\frac{1}{2}[2+(\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B})]-\cos ^2 \mathrm{C} \\
& =\frac{1}{2}\left[2+2 \cdot \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cdot \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\right]
\end{aligned}$
$-\cos ^2 \mathrm{C}$
$=1+\cos (A+B) \cdot \cos (A-B)-\cos 2 C$
In $\triangle \mathrm{ABC}$,
$\begin{aligned}
& A+B+C=\pi \\
& A+B=\pi-C \\
& \cos (A+B)=\cos (\pi-C) \\
& \cos (A+B)=-\cos C \ldots \ldots \ldots \ldots \ldots(i) \\
& \text { L.H.S. }=1-\cos C \cdot \cos (A-B)-\cos ^2 C
\end{aligned}$
$\ldots[$ From(i)]
$\begin{aligned}
& =1-\cos C \cdot[\cos (A-B)+\cos C] \\
& =1-\cos C \cdot[\cos (A-B)-\cos (A+B)]
\end{aligned}$
.. [From (i)]
$\begin{aligned}
& =1-\cos C .\left(2 . \sin A_{.} \sin B\right) \\
& =1-2. \sin A_{.} \sin B . \cos C \\
& =\text { R.H.S. }
\end{aligned}$
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Question 85 Marks
$\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2}=1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$
Answer
We know that, $\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}$
L.H.S.
$\begin{aligned}
&=\sin ^2 \frac{\mathrm{A}}{2}+\sin ^2 \frac{\mathrm{B}}{2}-\sin ^2 \frac{\mathrm{C}}{2} \\
&=\frac{1-\cos \mathrm{A}}{2}+\frac{1-\cos \mathrm{B}}{2}-\sin ^2 \frac{\mathrm{C}}{2} \\
&=\frac{1}{2}[2-(\cos \mathrm{A}+\cos \mathrm{B})]-\sin ^2 \frac{\mathrm{C}}{2} \\
&=\frac{1}{2}\left[2-2 \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right]-\sin ^2 \frac{\mathrm{C}}{2} \\
&=1-\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\sin ^2 \frac{\mathrm{C}}{2} \\
& \text { In } \Delta \mathrm{ABC}, \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\
& \therefore \quad \mathrm{A}+\mathrm{B}=\pi-\mathrm{C}\\
& \therefore \quad \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\cos \left(\frac{\pi-\mathrm{C}}{2}\right)=\cos \left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right) \\
& \quad=\sin \frac{\mathrm{C}}{2}.....(1)
\end{aligned}$
$\begin{aligned} & \therefore \quad \text { L.H.S. }=1-\sin \frac{\mathrm{C}}{2} \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\sin ^2 \frac{\mathrm{C}}{2} \\ & \text {...[From (i)] } \\ & =1-\sin \frac{C}{2} \cdot\left[\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)+\sin \frac{\mathrm{C}}{2}\right] \\ & =1-\sin \frac{C}{2} \cdot\left[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right] \\ & \ldots[\text { From (i)] } \\ & =1-\sin \frac{\mathrm{C}}{2} \cdot 2 \cos \left[\frac{\frac{\mathrm{A}+\mathrm{B}}{2}+\frac{\mathrm{A}-\mathrm{B}}{2}}{2}\right] \\ & \cos \left[\frac{\frac{A+B}{2}-\frac{A-B}{2}}{2}\right] \\ & =1-\sin \frac{\mathrm{C}}{2} \cdot\left(2 \cdot \cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2}\right) \\ & =1-2 \cdot \cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2} \cdot \sin \frac{\mathrm{C}}{2} \\ & =\text { R.H.S. } \\ & \end{aligned}$
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Question 95 Marks
$\begin{aligned}
& \cos A+\operatorname{Cos} B+C \cos C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\
& =2 \cdot \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)-\left(1-2 \sin ^2 \frac{C}{2}\right)
\end{aligned}$
Answer
$
\begin{aligned}
& \text { L.H.S. }=\sin A+\sin B+\sin C \\
& =2 \cdot \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)-\left(1-2 \sin ^2 \frac{C}{2}\right) \\
& \text { In } \triangle \mathrm{ABC}, \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \text { 。 } \\
& \therefore A+B=\pi-C \\
& \therefore \quad \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\cos \left(\frac{\pi-\mathrm{C}}{2}\right)=\cos \left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right) \\
& =\sin \frac{C}{2} \\
& \therefore \quad \text { L.H.S. }=2 \cdot \sin \frac{\mathrm{C}}{2} \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-1+2 \cdot \sin ^2 \frac{\mathrm{C}}{2} \\
& =2 \cdot \sin \frac{C}{2} \cdot\left[\cos \left(\frac{A-B}{2}\right)+\sin \frac{C}{2}\right]-1 \\
& =2 \cdot \sin \frac{C}{2} \cdot\left[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right]-1 \\
& =2 \sin \frac{\mathrm{C}}{2} \cdot 2 \cos \left[\frac{\frac{\mathrm{A}+\mathrm{B}}{2}+\frac{\mathrm{A}-\mathrm{B}}{2}}{2}\right] \\
\end{aligned}$
[From (i)]
..[From (i)]
$=2 \sin \frac{\mathrm{C}}{2} \cdot 2 \cos \left[\frac{\frac{\mathrm{A}+\mathrm{B}}{2}+\frac{\mathrm{A}-\mathrm{B}}{2}}{2}\right]$
$\begin{aligned}
& \left[\cdot \cos \left[\frac{\frac{A+B}{2}-\frac{A-B}{2}}{2}\right]-1\right. \\
& =2 \cdot \sin \frac{\mathrm{C}}{2} \cdot\left(2 \cdot \cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2}\right)-1 \\
& =4 \cdot \cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2} \cdot \sin \frac{\mathrm{C}}{2}-1 \\
& =\text { R.H.S. } \\
\end{aligned}$
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Question 105 Marks
In $\triangle A B C, A+B+C=\pi$ show that
$\sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Answer
$\begin{aligned} & \text { L.H.S. }=\sin A+\sin B+\sin C \\ & =2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cdot \cos \frac{C}{2} \\ & \text { In } \triangle A B C, A+B+C=\pi \\ & \therefore A+B=\pi-C \\ & \therefore \sin \left(\frac{A+B}{2}\right)=\sin \left(\frac{\pi-C}{2}\right)=\sin \left(\frac{\pi}{2}-\frac{C}{2}\right)=\cos \frac{C}{2} \ldots \text { (i) } \\ & \text { and } \cos \left(\frac{A+B}{2}\right)=\cos \left(\frac{\pi-C}{2}\right)=\cos \left(\frac{\pi}{2}-\frac{C}{2}\right)=\sin \frac{C}{2} \ldots \text { (ii) } \\ & \therefore \text { L.H.S. }=2 \cdot \cos \frac{C}{2} \cdot \cos \left(\frac{A-B}{2}\right)+2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \frac{C}{2} \ldots[F r o m(i) \\ & \text { and (ii)] } \\ & =2 \cdot \cos \frac{C}{2} \cdot\left[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right] \\ & =2 \cdot \cos \frac{C}{2} \cdot 2 \cos \left[\frac{A+\theta}{2}+\frac{A-B}{2}\right] \cdot \cos \left[\frac{A+B}{2}-\frac{A-\theta}{2}\right] \\ & =2 \cdot \cos \frac{C}{2} \cdot\left(2 \cos \frac{A}{2} \cdot \cos \frac{B}{2}\right) \\ & =4 \cdot \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} \\ & =R \cdot H \cdot S \cdot\end{aligned}$
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