Solve the Following Question.(3 Marks)

Question 13 Marks

Find the value of $\cos ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{4}{5}\right)$

Answer

$\text { Let } \cos ^{-1} \frac{3}{5}=\alpha \text { and } \cos ^{-1} \frac{4}{5}=\beta$
$\therefore \cos \alpha=\frac{3}{5} \text { and } \cos \beta=\frac{4}{5}$
$\therefore \sin \alpha=\sqrt{1-\cos ^2 \alpha}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
$\text { and } \sin \beta=\sqrt{1-\cos ^2 \beta}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Consider $\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
$=\frac{4}{5} \cdot \frac{4}{5}+\frac{3}{5} \cdot \frac{3}{5}$
$=\frac{16}{25}+\frac{9}{25}=\frac{25}{25}$
$=1$
$\therefore \alpha+\beta=\sin ^{-1}(1)$
$\text { i.e. } \cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{4}{5}=\frac{\pi}{2}$

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Question 23 Marks

In $\triangle ABC$ with usual notations, prove that : $2\left\{a \sin ^2\left(\frac{ C }{2}\right)+c \sin ^2\left(\frac{ A }{2}\right)\right\}=(a+c-b)$

Answer

$\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}$
L.H.S $=2\left\{a \sin ^2\left(\frac{C}{2}\right)+c \sin ^2\left(\frac{A}{2}\right)\right\}$
$=2\left\{\frac{a(1-\cos C)}{2}+\frac{c(1-\cos A)}{2}\right\}$
=a-acosC+c-ccosA
=(a+c)-(acosC+ccosA)
=a+c-b
R.H.S
$2 a \left\{\sin ^2\left(\frac{C}{2}\right)+c \sin ^2\left(\frac{A}{2}\right)\right\}=( a + c - b )$

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Question 33 Marks

Show that : $2 \sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{24}{7}\right)$

Answer

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Question 43 Marks

If $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$, then find the value of $x$.

Answer

$
\begin{array}{}
\text { As } 2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x) \\
\therefore \tan ^{-1}(\cos x)+\tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x) \\
\therefore \tan ^{-1}\left(\frac{\cos x+\cos x}{1-\cos ^2 x}\right)=\tan ^{-1}(2 \operatorname{cosec} x)\\
\therefore \quad \tan ^{-1}\left(\frac{2 \cos x}{\sin ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\
\therefore \quad \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \text {, but } \sin x \neq 0 \\
\therefore \quad \cos x=\sin x \\
\text { i.e. } \quad \frac{\cos x}{\sin x}=1 \\
\text { i.e. } \quad \cot x=1 \\
\therefore \quad x=\frac{\pi}{4} \\
\end{array}
$

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Question 53 Marks

Show that : $\sin ^{-1}\left(\frac{8}{17}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1}\left(\frac{77}{85}\right)$

Answer

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Question 63 Marks

Show that : $\sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{63}{16}\right)$

Answer

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Question 73 Marks

In $\triangle ABC$, prove that $\sin \left(\frac{ B - C }{2}\right)=\left(\frac{b-c}{a}\right) \cos \left(\frac{ A }{2}\right)$

Answer

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Question 83 Marks

Find the general solution of the equation $\sin 2 x+\sin 4 x+\sin 6 x=0$

Answer

$(\sin 2 x+\sin 6 x)+\sin 4 x=0$
$2 \sin 4 x \cdot \cos 2 x+\sin 4 x=0$
$\sin 4 x(2 \cos 2 x+1)=0$
$\sin 4 x=0$ or $2 \cos 2 x+1=0$
$\sin 4 x=0$ or $\cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)$
Using $\sin x=0 \Rightarrow x=n \pi$
$\sin 4 x=0$
$4 x=n \pi$
The genral solution is x
$x=\frac{n \pi}{4}$
using $\cos x=\cos \alpha \Rightarrow x=2 m x \pm \alpha$
$\cos 2 x=\cos \left(\frac{2 \pi}{3}\right)$
$2 x=2 \mp i \pm \frac{2 \pi}{3}$
The genral solution is $x$
$x=\mp i \pm \frac{\pi}{3}$ where $m, n \in z$

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Question 93 Marks

Find the general solution of $\cot x+\tan x=2 \operatorname{cosec} x$.

Answer

$\cot x+\tan x=2 \operatorname{coses} x$
$\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=\frac{2}{\sin x}$
$\frac{\cos ^2 x+\sin ^2 x}{\sin x \cdot \cos x}=\frac{2}{\sin x}$
$\left.\frac{1}{\sin x \cdot \cos x}=\frac{2}{\sin x} \ldots \ldots \ldots \ldots \ldots \ldots \ldots [\cos ^2 x+\sin ^2 x=1\right]$
$1=\frac{2}{\sin x} \times \sin x \cdot \cos x$
$1=2 \cos x$
$\cos x=\frac{1}{2}$
$x=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
Since, $\cos \theta=\cos \alpha$ implies $\theta=2 n \pi \pm \alpha, n \in Z$
the required general solution is $x=2 n \pi \pm \frac{\pi}{3}$ where $n \in Z$

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Question 103 Marks

If $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$, then find the value of ' $x$ '.

Answer

$\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$
$\tan ^{-1}\left(\frac{2 x+3 x}{1-(2 x)(3 x)}\right)=\frac{\pi}{4}$
$\therefore \frac{5 x}{1-6 x^2}=\tan \left(\frac{\pi}{4}\right)$
$\frac{5 x}{1-6 x^2}=1$
$5 x=1-6 x^2$
$6 x^2+5 x-1=0$
$6 x^2+6 x-x-1=0$
$6 x(x+1)-1(x+1)=0$
$(x+1)(6 x-1)=0$
$x=-1$ or $x=\frac{1}{6}$
But $x =-1$ does not satisfy $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$
$x=\frac{1}{6}$

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Question 113 Marks

In any triangle $\text{ABC}$ with usual notations prove $c=a \cos B +b \cos A$.

Answer

By cosine rule, we have $b^2 = c^2 + a^2 - 2ca \cos B$
$\therefore \cos B=\frac{c^2+a^2-b^2}{2} a c$
Similarly $\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\text{R.H.S.} = a \cos B + b \cos A$
$=a \cdot \frac{c^2+a^2-b^2}{2 a c}+b \cdot \frac{b^2+c^2-a^2}{2 b c}$
$=\frac{c^2+a^2-b^2}{2 c}+\frac{b^2+c^2-a^2}{2 c}$
$=\frac{c^2+a^2-b^2+b^2+c^2-a^2}{2 c}$
$=\frac{2 c^2}{2 c}$
$=c$
$=\text{L.H.s}$
$\therefore c = a \cos B + b \cos A$

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Question 123 Marks

Find the principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer

Let $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$\therefore \sin ^{-1} x=\left(\frac{1}{\sqrt{2}}\right)$
$\therefore \sin x=\sin \left(\frac{\pi}{4}\right)$
The principal value branch of
$\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\sin ^{-1}(\sin \theta)=\theta$
Hence, the required principal value of $x$ is $\frac{\pi}{4}$

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Maths Solve the Following Question.(3 Marks)

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