2 Marks Questions

Question 12 Marks

Find the value of $\cos \theta$ for which the equation $2 \cos \theta=x+\frac{1}{x}$ is possible, where $x$ is real.

Answer

Given equation is,
$2 \cos \theta=x+\frac{1}{x}$
$\Rightarrow \quad 2 \cos \theta=\frac{x^2+1}{x}$
$\begin{array}{l}\Rightarrow \quad x^2+1-2 x \cos \theta=0 \\ \Rightarrow \quad x^2-2 \cos \theta \cdot x+1=0\ldots\ldots (1)\end{array}$
Since, $x$ is real, so the discriminant must be greater than or equal to zero.
$\begin{aligned}4 \cos ^2 \theta-4 & \geq 0 \\\Rightarrow \quad 4\left(\cos ^2 \theta-1\right) & \geq 0 \quad \Rightarrow \cos ^2 \theta \geq 1\end{aligned}$
But $\cos ^2 \theta$ cannot be greater than 1 .
$\therefore \quad \cos ^2 \theta=1 \Rightarrow \cos \theta= \pm 1$

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Question 22 Marks

If $\sin \theta=\frac{12}{13}$ and $\theta$ is in second quadrant then find the value of $\sec \theta+\tan \theta$.

Answer

We know that
$\begin{array}{rlrl}\sin ^2 \theta+\cos ^2 \theta =1 \\\Rightarrow \quad \cos \theta= \pm \sqrt{1-\sin ^2 \theta}\end{array}$
$\cos \theta$ is negative is second quadrant.
$\therefore \quad \cos \theta=-\sqrt{1-\sin ^2 \theta}$
Now, $\sec \theta+\tan \theta=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=\frac{1+\sin \theta}{\cos \theta}$
$\begin{array}{l}=\frac{1+\sin \theta}{-\sqrt{1-\sin ^2 \theta}}=\frac{1+\frac{12}{13}}{-\sqrt{1-\left(\frac{12}{13}\right)^2}} \\=\frac{\frac{25}{13}}{-\sqrt{\frac{25}{169}}}=\frac{\frac{25}{13}}{-\frac{5}{13}}=-5\end{array}$

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Question 32 Marks

Prove that:
$
\sin ^2 \frac{\pi}{18}+\sin ^2 \frac{\pi}{9}+\sin ^2 \frac{7 \pi}{18}+\sin ^2 \frac{4 \pi}{9}=2
$

Answer

L.H.S $=\sin ^2 \frac{\pi}{18}+\sin ^2 \frac{\pi}{9}+\sin ^2 \frac{7 \pi}{18}+\sin ^2 \frac{4 \pi}{9}$
$\begin{array}{l}= \sin ^2 10^{\circ}+\sin ^2 20^{\circ}+\sin ^2 70^{\circ}+\sin ^2 80^{\circ} \\ = \sin ^2\left(90^{\circ}-80^{\circ}\right)+\sin ^2\left(90^{\circ}-70^{\circ}\right)+\sin ^2 70^{\circ} +\sin ^2 80^{\circ} \\ = \cos ^2 80^{\circ}+\cos ^2 70^{\circ}+\sin ^2 70^{\circ}+\sin ^2 80^{\circ} \\ {[\because \sin (90-\theta)=\cos \theta] } \\ =\left(\cos ^2 80^{\circ}+\sin ^2 80^{\circ}\right)+\left(\cos ^2 70^{\circ}+\sin ^2 70^{\circ}\right) \\ =1+1\end{array}$
$=2=\text {R.H.S.}$$\quad$ $\text {Hence proved}.$

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Question 42 Marks

Prove for any triangle :
$\begin{array}{l}\tan 2 A \cdot \tan 2 B \cdot \tan 2 C=\tan 2 A+\tan 2 B +\tan 2 C\end{array}$

Answer

$\because$ We know that :
$\begin{aligned}& A+B+C=180^{\circ} \\\Rightarrow \quad \therefore & 2 A+2 B+2 C=360^{\circ} \\\Rightarrow \quad & 2 A+2 B=360^{\circ}-2 C\end{aligned}$
(Taking tangent of both sides)
$\begin{array}{l}\Rightarrow \tan (2 A+2 B)=\tan \left(360^{\circ}-2 C\right) \\\Rightarrow \frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \cdot \tan 2 B}=-\tan 2 C \\\qquad \quad\left[\because \tan \left(180^{\circ}-\theta\right)=-\tan \theta\right] \\\Rightarrow \tan 2 A+\tan 2 B=-\tan 2 C \cdot(1-\tan 2 A \cdot \tan 2 B) \\\Rightarrow \tan 2 A+\tan 2 B=-\tan 2 C+\tan 2 A \cdot \tan 2 B \cdot \tan 2 C \end{array}$
$\Rightarrow \tan 2 A+\tan 2 B+\tan 2 C =\tan 2 A \cdot \tan 2 B \cdot \tan 2 C$
$ \begin{array}{l} \tan 2 A \cdot \tan 2 B \cdot \tan 2 C=\tan 2 A+\tan 2 B +\tan 2 C \text { Hence proved. } \end{array} $

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Question 52 Marks

Prove that:
$
\frac{1+\sin 2 \theta-\cos 2 \theta}{1+\sin 2 \theta+\cos 2 \theta}=\tan \theta
$

Answer

L.H.S.
$\begin{array}{l}=\frac{1+\sin 2 \theta-\cos 2 \theta}{1+\sin 2 \theta+\cos 2 \theta} \\=\frac{(1-\cos 2 \theta)+\sin 2 \theta}{(1+\cos 2 \theta)+\sin 2 \theta} \\=\frac{2 \sin ^2 \theta+2 \sin \theta \cos \theta}{2 \cos ^2 \theta+2 \sin \theta \cos \theta} \\=\frac{2 \sin \theta(\sin \theta+\cos \theta)}{2 \cos \theta(\sin \theta+\cos \theta)} \\\end{array}$
$=\frac{\sin \theta}{\cos \theta}=\tan \theta=$ R.H.S. $\quad$ Hence proved.

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Question 62 Marks

Prove that:
$
\begin{array}{l}
\sin (n+1) A \sin (n+2) A+\cos (n+1) A \cos (n+2) A=\cos A
\end{array}
$

Answer

$\begin{array}{l}\text { L.H.S. }=\sin (n+1) A \sin (n+2) A+\cos (n+1) \\A \cos (n+2) A \\\Rightarrow \quad \cos (n+2) A \cos (n+1) A+\sin (n+2) A \sin (n+1) A \\\Rightarrow \quad \cos [(n+2) A-(n+1) A]= \cos [(n+2-n-1) A] \\=\cos A=RHS\end{array}$

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Question 72 Marks

Prove that:
$
\sec \left(\frac{\pi}{4}+\theta\right) \sec \left(\frac{\pi}{4}-\theta\right)=2 \sec 2 \theta
$

Answer

$\begin{array}{l}\text { L.H.S. } \sec \left(\frac{\pi}{4}+\theta\right) \cdot \sec \left(\frac{\pi}{4}-\theta\right) \\ \Rightarrow \frac{1}{\cos \left(\frac{\pi}{4}+\theta\right) \cos \left(\frac{\pi}{4}-\theta\right)}=\frac{2}{2 \cos \left(\frac{\pi}{4}+\theta\right) \cos \left(\frac{\pi}{4}-\theta\right)} \\ \Rightarrow \frac{2}{\cos \left(\frac{\pi}{4}+\theta+\frac{\pi}{4}-\theta\right)+\cos \left(\frac{\pi}{4}+\theta-\frac{\pi}{4}+\theta\right)}\end{array}$
$[\because 2 \cos A \cos B=\cos (A+B)-\cos (A-B)]$
$\Rightarrow \frac{2}{\cos \frac{\pi}{2}+\cos 2 \theta}=\frac{2}{0+\cos 2 \theta}$
$\Rightarrow \frac{2}{\cos 2 \theta}=2 \sec 2 \theta=$ R.H.S.

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Question 82 Marks

If $\tan \theta=\frac{a}{b}$, prove that :
$
b \cos 2 \theta+a \sin 2 \theta=b
$

Answer

Given, $\tan \theta=\frac{a}{b}$
Now, L.H.S. $=b \cos 2 \theta+a \sin 2 \theta$
Using the formula
$=b \cdot \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}+a \cdot \frac{2 \tan \theta}{1+\tan ^2 \theta}$
Putting the value
$=b \cdot \frac{1-\left(\frac{a}{b}\right)^2}{1+\left(\frac{a}{b}\right)^2}+2 a \cdot \frac{\frac{a}{b}}{1+\left(\frac{a}{b}\right)^2}$
$=b \cdot \frac{b^2-a^2}{b^2+a^2}+\frac{2 a^2 b}{b^2+a^2}$
$=\frac{b\left(b^2-a^2+2 a^2\right)}{\left(b^2+a^2\right)}=\frac{b\left(b^2+a^2\right)}{\left(b^2+a^2\right)}=b=$ R.H.S.

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Question 92 Marks

If $A+B=45^{\circ}$, then prove that :
$(\cot A-1)(\cot B-1)=2$

Answer

$\text {Given,}\quad A+B=45^{\circ} $
$\Rightarrow \quad \cot ( A + B )=\cot 45^{\circ}=1$
Now, $\quad \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}$
$\Rightarrow \quad 1=\frac{\cot A \cot B -1}{\cot A+\cot B }$
$\begin{array}{l}\Rightarrow \quad \cot A +\cot B =\cot A \cot B -1 \\ \Rightarrow \quad \cot A \cot B -\cot A -\cot B =1\end{array}$
Adding 1 on both sides :
$\begin{array}{l}\Rightarrow \quad 2=\cot A \cot B-\cot A-\cot B+1 \\\Rightarrow \quad 2=\cot A(\cot B-1)-1(\cot B-1) \\\Rightarrow \quad 2=(\cot A-1)(\cot B-1) \\\Rightarrow \quad(\cot A-1)(\cot B-1)=2\end{array}$

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Question 102 Marks

Prove the following for quadrilateral ABCD :
(i) $\sin (A+B)+\sin (C+D)=0$
(ii) $\cos ( A + B )=\cos ( C + D )$

Answer

(i) We know that in any quadrilateral,
$A+B+C+D=2 \pi$
$\begin{array}{ll}\Rightarrow & A + B =2 \pi-( C + D ) \\ \Rightarrow & \sin ( A + B )=\sin [2 \pi-( C + D )] \\ \Rightarrow & \sin ( A + B )=-\sin ( C + D ) \\ & {[\because \sin (2 \pi-\theta)=-\sin \theta]}\end{array}$
$ \Rightarrow \sin (A+B)+\sin (C+D)=0 $
(ii) We know that
$A + B + C + D =2 \pi$
$\Rightarrow \quad( A + B )=2 \pi-( C + D )$
$\Rightarrow \quad \cos ( A + B )=\cos [2 \pi-( C + D )]$
$\begin{array}{l} \Rightarrow \quad \cos (A+B)=\cos (C+D) \\ \quad\quad{[\because \cos (2 \pi-\theta)=\cos \theta] }\end{array}$

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Question 112 Marks

Prove that :
$\frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}=0$

Answer

Solving first term of L.H.S.
$\begin{array}{l}\frac{\sin (A-B)}{\cos A \cos B}=\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B} \\=\frac{\sin A \cos B}{\cos A \cos B}-\frac{\cos A \sin B}{\cos A \cos B} \\=\tan A-\tan B\end{array}$
Similarly second term $=\tan B-\tan C$ and third term $=\tan C -\tan A$
Adding the values of all three terms:
$\begin{array}{l}\tan A-\tan B+\tan B-\tan C+\tan C-\tan A \\\Rightarrow \quad 0=\text { R.H.S. }\end{array}$

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Question 122 Marks

Find the value of $\tan 105^{\circ}$.

Answer

$\begin{array}{l}\tan 105^{\circ}=\tan \left(60^{\circ}+45^{\circ}\right)=\frac{\tan 60^{\circ}+\tan 45^{\circ}}{1-\tan 60^{\circ} \tan 45^{\circ}} \\ \begin{aligned} \tan 105^{\circ} & =\frac{\sqrt{3}+1}{1-\sqrt{3} \times 1}=\frac{\sqrt{3}+1}{1-\sqrt{3}} \\ \tan 105^{\circ} & =\frac{\sqrt{3}+1}{1-\sqrt{3}} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{(\sqrt{3}+1)^2}{1-3} \\ & =\frac{(\sqrt{3})^2+2 \times \sqrt{3}+(1)^2}{-2} \\ & =\frac{3+2 \sqrt{3}+1}{-2}=\frac{4+2 \sqrt{3}}{-2} \\ & =\frac{2(2+\sqrt{3})}{-2}=-(2+\sqrt{3})\end{aligned}\end{array}$

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