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Maths Solve the following Question.(1 Marks)

Trigonometry – II · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 33 questions.

Questions

Solve the following Question.(1 Marks)

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

Question 11 Mark
$\sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4}$
Answer
We know that, $\sin ^2 \theta=1-\cos ^2 \theta$
$\begin{aligned}
& \sin ^2 36^{\circ}=1-\cos ^2 36^{\circ} \\
& =1-\left(\frac{\sqrt{5}+1}{4}\right)^2 \\
& =\frac{16-(5+1+2 \sqrt{5})}{16} \\
& =\frac{10-2 \sqrt{5}}{16} \\
& \therefore \sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4} \ldots \ldots\left[\because \sin 36^{\circ} \text { is positive }\right]
\end{aligned}$
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Question 41 Mark
Prove that $\frac{\cos ^3 \theta-\cos 3 \theta}{\cos \theta}+\frac{\sin ^3 \theta+\sin 3 \theta}{\sin \theta}=3$
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Question 71 Mark
Prove the following : $\sin \frac{\pi}{15}+\sin \frac{4 \pi}{15}-\sin \frac{14 \pi}{15}-\sin \frac{11 \pi}{15}=0$
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Question 91 Mark
Express the following as sum or difference of two trigonometric function : $4 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$
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Question 101 Mark
Express the following as sum or difference of two trigonometric function: $2 \sin 4 \theta \cos 2 \theta$
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Question 151 Mark
Express the following as a sum or difference of two trigonometric functions : $2 \cos 35^{\circ} \cos 75^{\circ}$
Answer
$2 \cos 35^{\circ} \cos 75^{\circ}$
$\begin{aligned}
& =\cos \left(35^{\circ}+75^{\circ}\right)+\cos \left(35^{\circ}-75^{\circ}\right) \\
& =\cos 110^{\circ}+\cos (-40)^{\circ} \\
& =\cos 110^{\circ}+\cos 40^{\circ} \ldots[\because \cos (-\theta)=\cos \theta]
\end{aligned}$
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Question 161 Mark
Express the following as a sum or difference of two trigonometric functions : $2 \cos 4 \theta \cos 2 \theta$
Answer
$\begin{aligned}
& \text2 \cos 4 \theta \cos 2 \theta=\cos (4 \theta+2 \theta)+\cos (4 \theta-2 \theta) \\
& =\cos 6 \theta+\cos 2 \theta
\end{aligned}$
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Question 171 Mark
Express the following as a sum or difference of two trigonometric functions : $2 \sin \frac{2 \pi}{3} \cos \frac{\pi}{2}$
Answer
$\begin{aligned}
2 \sin \frac{2 \pi}{3} \cos \frac{\pi}{2} & =\sin \left(\frac{2 \pi}{3}+\frac{\pi}{2}\right)+\sin \left(\frac{2 \pi}{3}-\frac{\pi}{2}\right) \\
& =\sin \frac{7 \pi}{6}+\sin \frac{\pi}{6}
\end{aligned}
$
[Note: Answer given in the textbook is $\sin \frac{7 \pi}{12}+\sin \frac{\pi}{12}$ However, as per our calculation it is $\sin \frac{7 \pi}{6}+\sin \frac{\pi}{6}$
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Question 181 Mark
Express the following as a sum or difference of two trigonometric functions : $2 \sin 4 x \cos 2 x$
Answer
$\begin{aligned}
& \text { i. } 2 \sin 4 x \cos 2 x=\sin (4 x+2 x)+\sin (4 x-2 x) \\
& =\sin 6 x+\sin 2 x
\end{aligned}$
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Question 191 Mark
Prove the following : $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^2 x$
Answer
$
\begin{aligned}
\text { L.H.S. } & =\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)} \\
& =\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \\
& =\frac{\cos ^2 x}{\sin ^2 x} \\
& =\cot ^2 x \\
& =\text { R.H.S. }
\end{aligned}
$
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Question 201 Mark
Find the values of : cot (-1110°)
Answer
\begin{aligned}
& \cot \left(-1110^{\circ}\right)=-\cot \left(1110^{\circ}\right) \\
& =-\cot \left(1080^{\circ}+30^{\circ}\right) \\
& =-\cot \left(3 \times 360^{\circ}+30^{\circ}\right) \\
& =-\cot 30^{\circ} \\
& =-\sqrt{3} \\
& =-1,
\end{aligned}
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Question 211 Mark
Find the values of : cosec 780°
Answer
\begin{aligned}
& \operatorname{cosec} 780^{\circ}=\operatorname{cosec}\left(720^{\circ}+60^{\circ}\right) \\
& =\operatorname{cosec}\left(2 \times 360^{\circ}+60^{\circ}\right) \\
& =\operatorname{cosec} 60^{\circ} \\
& =\frac{2}{\sqrt{3}} \\
\end{aligned}
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Question 221 Mark
Find the values of : sec (-855°)
Answer
\begin{aligned}
& \sec \left(-855^{\circ}\right)=\sec \left(855^{\circ}\right) \\
& =\sec \left(720^{\circ}+135^{\circ}\right) \\
& =\sec \left(2 \times 360^{\circ}+135^{\circ}\right)=\sec 135^{\circ} \\
& =\sec \left(90^{\circ}+45^{\circ}\right) \\
& =-\operatorname{cosec} 45^{\circ} \\
& =-\sqrt{2} \\
\end{aligned}
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Question 231 Mark
Find the values of : sec 240°
Answer
\begin{aligned}
& \sec 240^{\circ}=\sec \left(180^{\circ}+60^{\circ}\right) \\
& =-\sec 60^{\circ} \\
& =-2 \\
\end{aligned}
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Question 241 Mark
Find the values of : tan (-690°)
Answer
\begin{aligned}
& \tan \left(-690^{\circ}\right)=-\tan 690^{\circ} \\
& =-\tan \left(720^{\circ}-30^{\circ}\right) \\
& =-\tan \left(2 \times 360^{\circ}-30^{\circ}\right) \\
& =-\left(-\tan 30^{\circ}\right) \\
& =\tan 30^{\circ} \\
& =\frac{1}{\sqrt{3}} \\
\end{aligned}
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Question 251 Mark
Find the values of : tan 225°
Answer
$
\begin{aligned}
& \tan 225^{\circ}=\tan \left(180^{\circ}+45^{\circ}\right) \\
& =\tan 45^{\circ} \\
& =1.
\end{aligned}
$
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Question 261 Mark
Find the values of : cos 600°
Answer
$
\begin{aligned}
& \cos 600^{\circ}=\cos \left(360^{\circ}+240^{\circ}\right) \\
& =\cos 240^{\circ} \\
& =\cos \left(180^{\circ}+60^{\circ}\right) \\
& =-\cos 60^{\circ} \\
& =-\frac{1}{2}
\end{aligned}
$
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Question 271 Mark
Find the values of : cos 315°
Answer
$
\begin{aligned}
& \cos 315^{\circ}=\cos \left(270^{\circ}+45^{\circ}\right) \\
& \sin 45^{\circ}=\frac{1}{\sqrt{2}}
\end{aligned}
$
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Question 281 Mark
Find the values of : sin 495°
Answer
$
\begin{aligned}
& \sin 495^{\circ}=\sin \left(360^{\circ}+135^{\circ}\right) \\
& =\sin \left(135^{\circ}\right) \\
& =\sin \left(90^{\circ}+45^{\circ}\right) \\
& =\cos 45^{\circ} \\
& =\frac{1}{\sqrt{2}} \\
\end{aligned}
$
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Question 291 Mark
Find the values of : sin 690°
Answer
$
\begin{aligned}
& \sin 690^{\circ}=\sin \left(720^{\circ}-30^{\circ}\right) \\
& =\sin \left(2 \times 360^{\circ}-30^{\circ}\right) \\
& =-\sin 30^{\circ} \\
& =\frac{-1}{2}
\end{aligned}
$
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Question 301 Mark
Find the values of : cot 225°
Answer
$\begin{aligned} \cot 225^{\circ} & =\frac{1}{\tan 225^{\circ}}=\frac{1}{\tan \left(180^{\circ}+45^{\circ}\right)} \\ & =\frac{1}{\left(\frac{\tan 180^{\circ}+\tan 45^{\circ}}{1-\tan 180^{\circ} \tan 45^{\circ}}\right)} \\ & =\frac{1}{\left(\frac{0+1}{1-0(1)}\right)} \\ & =\frac{1}{\left(\frac{1}{1}\right)} \\ & =1\end{aligned}$
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Question 311 Mark
Find the values of : tan 105°
Answer
$\begin{aligned} & =\frac{\tan 60^{\circ}+\tan 45^{\circ}}{1-\tan 60^{\circ} \tan 45^{\circ}} \\ & =\frac{\sqrt{3}+1}{1-(\sqrt{3})(1)}\end{aligned}$
$\begin{aligned} & =\frac{\sqrt{3}+1}{1-\sqrt{3}}\end{aligned}$
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Question 321 Mark
Find the values of : cos 75°
Answer
$\begin{aligned} & \cos 75^{\circ}=\cos \left(45^{\circ}+30^{\circ}\right) \\ & =\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ} \\ & =\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \\ & =\frac{\sqrt{3}-1}{2 \sqrt{2}}\end{aligned}$
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Question 331 Mark
Find the values of : sin 150°
Answer
$\begin{aligned}
& \sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right) \\
& =\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\
& \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}}
\end{aligned}$
[Note: Answer given in the textbook is $\frac{\sqrt{3}+1}{2 \sqrt{2}}$ However, as per our calculation it is $\frac{\sqrt{3}-1}{2 \sqrt{2}}$
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