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RD Sharma Class 12 Exercise 3.13 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 3.13 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.13 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Updated on 21 Jan 2022, 11:46 AM IST

The RD Sharma solution books are the best companion for the class 12 students. When a teacher is the best guide at the school to clear the doubts, the RD Sharma solution books are the best guides for the students at their home. Many students find it challenging to cope with the concepts in the 3rd chapter of mathematics, Inverse Trigonometry. The RD Sharma Class 12th Exercise 3.13 books are a boon for them.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

Inverse Trigonometry is not an easy chapter to solve, and there are numerous formulas and methods to bring the right solutions. Especially in the thirteenth exercise, it is even more challenging for students to solve when they are unclear about the previous exercises. Exercise 3.13 consists of problems that include the concepts like sine, cosine, tangent, cotangent, secant, and cosecant functions. There are five questions, including its subparts, to be solved in this exercise. To make it easier for the students, the RD Sharma Class 12 Chapter 3 Exercise 3.13 will lend a helping hand to the students.

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometry Function Ex 3.13

Inverse Trigonometric Function Exercise 3.13 Question 1

Answer:

Given:
$\cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}= \alpha$
To prove:
$9x^{2}-12xy\cos \alpha +4y^{2}= 36\sin ^{2}\alpha$
Hint:
We are applying the formula of
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right )\right ]$
Solution:
We have
$\cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}= \cos^{-1}\left [ \frac{x}{2}\times \frac{y}{3}-\left ( \sqrt{1-\left ( \frac{x}{2} \right )^{2}} \right )\left ( \sqrt{1-\left ( \frac{y}{3} \right )^{2}} \right )\right ]$
$= \cos^{-1}\left [ \frac{xy}{6}-\frac{\sqrt{4-x^{2}}}{2}\times \frac{\sqrt{9-y^{2}}}{3} \right ]$
$= \cos^{-1}\left [ \frac{xy-\sqrt{4-x^{2}}\times \sqrt{9-y^{2}}}{6} \right ]= \alpha$ (Let)
$\Rightarrow xy-\sqrt{4-x^{2}} \times \sqrt{9-y^{2}} = 6\cos \alpha$
$\Rightarrow \left ( xy-6\cos \alpha \right )= \sqrt{4-x^{2}}\times \sqrt{9-y^{2}}$
On squaring both sides, we get
$\! \! \! \! \! \! \! \! \Rightarrow \left ( xy-6\cos \alpha \right )^{2}= \left ( 4-x^{2} \right )\left ( 9-y ^{2}\right )\\\Rightarrow x^{2y^{2}}+36\cos ^{2}\alpha -12xy\cos \alpha = 36-9x^{2}-4y^{2}+x^{2}y^{2}$
$\! \! \! \! \! \! \! \! \! \Rightarrow 9x^{2}+4y^{2}-36+36\cos ^{2}\alpha-12xy\cos \alpha = 0\\\Rightarrow 9x^{2}+4y^{2}-12xy\cos \alpha-36\left ( 1-\cos ^{2}\alpha \right )= 0$
$\Rightarrow 9x^{2}+4y^{2}-12xy\cos \alpha -36\sin ^{2}\alpha = 0$
$\Rightarrow 9x^{2}+4y^{2}-12xy\cos \alpha = 36\sin ^{2}\alpha$
Hence proved

Inverse Trigonometric Function Exercise 3.13 Question 2

Answer:
$x= ab$
Given:
$\cos^{-1}\frac{a}{x}-\cos^{-1}\frac{b}{x}= \cos^{-1}\frac{1}{b}-\cos^{-1}\frac{1}{a}$
Hint:
First, we will separate the angle of $a$ terms then we will apply the formula,
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right ) \right ]$
$\cos^{-1}\frac{a}{x}+\cos^{-1}\frac{1}{a} = \cos^{-1}\frac{1}{b}+\cos^{-1}\frac{b}{x}$
Solution:
We know that,
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right ) \right ]$
Substituting the values in the formula, we get
$\cos^{-1}\left [ \frac{1}{x}-\sqrt{1-\left ( \frac{a}{x} \right )^{2}} \sqrt{1-\left ( \frac{1}{a} \right )^{2}}\right ]= \cos^{-1}\left [ \frac{1}{x}\sqrt{1-\left ( \frac{b}{x} \right )^{2}}\sqrt{1-\left ( \frac{1}{b} \right )^{2}} \right ]$
$\Rightarrow \frac{1}{x}-\sqrt{1-\left ( \frac{a}{x} \right )^{2}} \sqrt{1-\left ( \frac{1}{a} \right )^{2}}=\frac{1}{x}-\sqrt{1-\left ( \frac{b}{x} \right )^{2}}\sqrt{1-\left ( \frac{1}{b} \right )^{2}}$
$\Rightarrow\sqrt{1-\left ( \frac{a}{x} \right )^{2}} \sqrt{1-\left ( \frac{1}{a} \right )^{2}}=\sqrt{1-\left ( \frac{b}{x} \right )^{2}}\sqrt{1-\left ( \frac{1}{b} \right )^{2}}$
Squaring on both sides, we get
$\Rightarrow \left ( 1-\left ( \frac{a}{x} \right )^{2}\right )\left ( 1-\left ( \frac{1}{a} \right )^{2} \right )=\left ( 1-\left ( \frac{b}{x} \right ) ^{2}\right )\left ( 1-\left ( \frac{1}{b}\right )^{2} \right )$
$\Rightarrow 1-\left ( \frac{a}{x} \right )^{2}-\left ( \frac{1}{a} \right )^{2}+\left ( \frac{1}{x} \right )^{2}= 1-\left ( \frac{b}{x} \right )^{2}-\left ( \frac{1}{b} \right )^{2}+\left ( \frac{1}{x} \right )^{2}$
$\Rightarrow \left ( \frac{b}{x} \right )^{2}-\left ( \frac{a}{x} \right )^{2}= \left ( \frac{1}{a} \right )^{2}-\left ( \frac{1}{b} \right )^{2}$
On Simplifying, we get
$\Rightarrow \left ( b^{2} -a^{2}\right )a^{2}b^{2}= x^{2}\left ( b^{2}-a^{2} \right )$
$\Rightarrow x^{2}= a^{2}b^{2}$
$\Rightarrow x= ab$
Hence the result.

Inverse Trigonometric Function Exercise 3.13 Question 3

Answer:
$x= \frac{1}{2}$
Given:
$\cos^{-1}\sqrt{3x}+\cos^{-1}x= \frac{x}{2}$
Hint:
We are applying the formula,
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right ) \right ]$
Solution:
We have
$\cos^{-1}\sqrt{3x}+\cos^{-1}x= \frac{x}{2}$
Using formula written on the hint, we get
$\Rightarrow \cos^{-1}\left [ \sqrt{3x}\times x-\left ( \sqrt{1-\left ( \sqrt{3x} \right )^{2}} \right )\left ( \sqrt{1-x^{2}} \right ) \right ]= \frac{x}{2}$
$\Rightarrow \cos^{-1}\left [ \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right ) \right ]= \frac{x}{2}$
$\Rightarrow \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right )= \cos \frac{x}{2}$
$\Rightarrow \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right )= 0\; \; \; \; \; \; \left [ \because \cos \frac{x}{2}= 0 \right ]$
$\Rightarrow \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right )$
Squaring on both sides, we get
$\Rightarrow 3x^{4}-\left ( 1-3x ^{2} \right )\left ( 1-x^{2} \right )$
$\Rightarrow 3x^{4}= 1-x ^{2} -3x^{2}+3x^{4}$
$\Rightarrow 1-4x^{2}= 0$
$\Rightarrow 4x^{2}= 1$
$\Rightarrow x^{2}= \frac{1}{4}$
$\Rightarrow x=\pm \frac{1}{2}\; \; \; \; \; \; \; \; \; [ As -\frac{1}{2} is\: not\: satisf\! ying \: the\: equation]$
Hence the result is $x= \frac{1}{2}$

Inverse Trigonometric Function Exercise 3.13 Question 4

Answer:

Given:
$\cos^{-1}\left ( \frac{4}{5} \right )+\cos^{-1}\left ( \frac{12}{13} \right )$
To prove:
$\cos^{-1}\left ( \frac{4}{5} \right )+\cos^{-1}\left ( \frac{12}{13} \right )= \cos^{-1}\left ( \frac{33}{65} \right )$
Hint:
We will use the formula on L.H.S
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right ) \right ]$
Solution:
Taking L.H.S
L.H.S
$\cos^{-1}\left ( \frac{4}{5} \right )+\cos^{-1}\left ( \frac{12}{13} \right )$

$\Rightarrow \cos^{-1}\left [ \frac{48}{65}-\sqrt{1-\frac{16}{25} }\sqrt{1-\frac{144}{169}}\right ]$
$\Rightarrow \cos^{-1}\left [ \frac{48}{65}-\frac{3}{5}\times \frac{5}{13}\right ]$
$\Rightarrow \cos^{-1}\left ( \frac{33}{65} \right )$
$=$ R.H.S
Hence we get R.H.S.

Inverse Trigonometric Function Exercise 3.13 Question 5

Answer:

Given:
$\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}$
Hint:
First, we convert $\cos^{-1}$ to $\sin^{-1}$ and then use $\sin \left ( a+b \right )$ formula.
Solution:
Let $a= \cos^{-1}\frac{12}{13}$ and $b= \sin^{-1}\frac{3}{5}$
Finding $\sin a , \cos a$.
Let $a= \cos^{-1}\frac{12}{13}$
$\cos a= \left (\frac{12}{13} \right )$
We know that
$\sin a= \sqrt{1-\cos ^{2}}a$
$= \sqrt{1-\left ( \frac{12}{13} \right )^{2}}$
$= \sqrt{\frac{25}{169}}$
$= \frac{5}{13}$
Again, finding $\sin b, \cos b$
Let $b= \sin^{-1}\left ( \frac{3}{5} \right )$
$\Rightarrow \sin b= \frac{3}{5}$
We know that,
$\cos b= \sqrt{1-\sin ^{2}b}$
$\Rightarrow \cos b= \sqrt{1-\left ( \frac{3}{5} \right )^{2}}$
$= \sqrt{\frac{16}{25}}= \frac{4}{5}$
$\cos b= \frac{4}{5}$
We know that,
$\sin \left ( a+b \right )= \sin a\cos b+\cos a\sin b$
Putting, $\sin a= \frac{5}{13},\cos a= \frac{12}{13}$
$\sin b= \frac{3}{5},\cos b= \frac{4}{5}$
We have,
$\sin \left ( a+b \right )= \frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}$
$= \frac{20}{65}+\frac{36}{65}$
$\sin \left ( a+b \right )= \frac{56}{65}$
$a+b =\sin^{-1} \frac{56}{65}$
Putting the value of $a,b$
$\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}= \sin^{-1}\frac{56}{65}$
LHS=RHS
Hence proved.

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