### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

### Inverse Trigonometric Functions Chapter 2 Class 12

INVERSE   TRIGONOMETRIC   FUNCTIONS
Maths formulas based on Inverse trigonometric functions chapter 2 class 12, useful and important maths formulas on Inverse Trigonometric Functions

Note: Click here to open the Assignment on Inverse Trigonometric Functions Chapter 2 Class 12

Before we start the Inverse Trigonometric functions we should know the trigonometric functions class 11 chapter 3. First of all students should learn all the formulas of trigonometric functions then start this chapter.

Inverse Trigonometric Functions
CBSE Syllabus For Inverse Trigonometric Functions Chapter 2 Class XII

 Definition, range, domain, principal value branch, Graphs of inverse trigonometric functions, Elementary Properties of Inverse Trigonometric Functions

1.) In this section we discuss about the principal value
2.) Domain, and range of Inverse Trigonometric Functions
3.) Properties of Inverse Trigonometric Functions.
4.) Graph of Inverse of Trigonometric Functions.
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Simple meanings of Inverse of a Function

$y=sin^{-1}x\: \: \Rightarrow \: \: siny=x$

$f^{-1}of(x)\: \: \Rightarrow \: \: f^{-1}\left [ f(x) \right ]=x$

$sin^{-1}\:o \:\: sin\: x\Rightarrow \: \: sin^{-1}\left ( sin\:x \right )=x$

$sin^{2}x=(sinx)^{2} \: \: \: but \: \: \: sin^{-1}x\neq (sinx)^{-1}$

sinx is a function and sin-1x  is an angle

 Functions Domain  (Value of x) Range (Value of y) Principal Value Branch $y=sin^{-1}x$ [-1,1] $\left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]$ $y=cos^{-1}x$ [-1,1] $[0,\pi ]$ $y=tan^{-1}x$ R $\left (-\frac{\pi }{2},\frac{\pi }{2} \right )$ $y=cot^{-1}x$ R $(0,\pi )$ $y=sec^{-1}x$ R-(-1,1) $[0,\pi ]-\left \{ \frac{\pi }{2} \right \}$ $y=cosec^{-1}x$ R-(-1,1) $\left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]-\left \{ 0 \right \}$

 Functions Principal Value Branch $y=sin^{-1}x$ $-\frac{\pi }{2}\leq y\leq \frac{\pi }{2}$ $y=cos^{-1}x$ $0\leq y\leq \pi$ $y=tan^{-1}x$ $-\frac{\pi }{2}< y< \frac{\pi }{2}$ $y=cot^{-1}x$ $0< y< \pi$ $y=sec^{-1}x$ $0\leq y\leq \pi ,\; \; y\neq \frac{\pi }{2}$ $y=cosec^{-1}x$ $-\frac{\pi }{2}\leq y\leq \frac{\pi }{2}, \; \; y\neq 0$

$sin(sin^{-1}x)=x,\: \: \: \: x\: \epsilon\: [-1,1]$

$cos(cos^{-1}x)=x,\: \: \: \: x\: \epsilon\: [-1,1]$

$tan(tan^{-1}x)=x,\: \: \: \: x\: \epsilon\: R$

$cot(cot^{-1}x)=x,\: \: \: \: x\: \epsilon\: R$

$sec(sec^{-1}x)=x,\: \: \: \: x\: \epsilon\: R-(-1,1)$

$cosec(cosec^{-1}x)=x,\: \: \: \: x\: \epsilon\: R-(-1,1)$

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Important Note:

$sin^{-1}\left ( \frac{P}{H} \right )=cos^{-1}\left (\frac{B}{H} \right )=tan^{-1}\left ( \frac{P}{B} \right )$

$cot^{-1}\left ( \frac{B}{P} \right )=sec^{-1}\left ( \frac{H}{B} \right )=cosec^{-1}\left ( \frac{H}{P} \right )$

Principal Value:-

The value of the inverse trigonometric functions which lies in its principal value branch is called its Principal Value.

Inverse Trigonometric Functions with Negative angles

$sin^{-1}(-x)=-sin^{-1}x$

$cos^{-1}(-x)=\pi -cos^{-1}x$

$tan^{-1}(-x)=-tan^{-1}x$

$cot^{-1}(-x)=\pi -cot^{-1}x$

$sec^{-1}(-x)=\pi -sec^{-1}x$

$cosec^{-1}(-x)=-cosec^{-1}x$

Properties of inverse trigonometric functions:-

$sin^{-1}x + cos^{-1}x=\frac{\pi }{2}\: \: \: \Rightarrow \: \: sin^{-1}x=\frac{\pi }{2}-cos^{-1}x$

$tan^{-1}x + cot^{-1}x=\frac{\pi }{2}\: \: \: \Rightarrow \: \: tan^{-1}x=\frac{\pi }{2}-cot^{-1}x$

$sec^{-1}x + cosec^{-1}x=\frac{\pi }{2}\: \: \: \Rightarrow \: \: sec^{-1}x=\frac{\pi }{2}-cosec^{-1}x$
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$sin^{-1}\frac{1}{x}= cosec^{-1}x\: \: ,\: \: cos^{-1}\frac{1}{x}=sec^{-1}x\: ,\: \: tan^{-1}\frac{1}{x}=cot^{-1}x$
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$2tan^{-1}x=sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )=cos^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right )=tan^{-1}\left ( \frac{2x}{1-x^{2}} \right )$
Other Important Results

$sin^{-1}x+sin^{-1}y=sin^{-1}\left [ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} \right ]$

$sin^{-1}x-sin^{-1}y=sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$

$cos^{-1}x+cos^{-1}y=cos^{-1}\left [ xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}} \right ]$

$cos^{-1}x-cos^{-1}y=cos^{-1}\left [ xy+\sqrt{1-x^{2}}\sqrt{1-y^{2}} \right ]$

$tan^{-1}x+tan^{-1}y=tan^{-1}\left ( \frac{x+y}{1-xy} \right )$

$tan^{-1}x-tan^{-1}y=tan^{-1}\left ( \frac{x-y}{1+xy} \right )$

$tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}\left ( \frac{x+y+z-xyz}{1-xy-yz-zx} \right )$
SUBSTITUTION PROPERTIES :-

For a2 + b2, substitution is x = atan θ or x = acot θ

For a2 - b2, substitution is x = asin θ or x = acot θ

For x2 - a2, substitution is x = asec θ or x = acosec θ

$For\: \sqrt{\frac{a+x}{a-x}}\: \: or\: \: \sqrt{\frac{a-x}{a+x}}$  the substitution is x = acos2θ

$For\: \sqrt{\frac{a^{2}+x^{2}}{a^{2}-x^{2}}}\: \: or\: \: \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$

$For\: \sqrt{\frac{a^{2}+x^{2}}{a^{2}-x^{2}}}\: \: or\: \: \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$ the substitution is x2 = a2cos2θ

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Chapter 2 Miscellaneous Exercise Q 14

Prove that:  $tan^{-1}\left (\frac{1-x}{1+x} \right )=\frac{1}{2}tan^{-1}x$

Solution

$Let\; \; x=tan\theta \; \; \; \Rightarrow \; \; \theta =tan^{-1}x\$

$tan^{-1}\left ( \frac{1-tan\theta }{1+tan\theta } \right )=\frac{1}{2}tan^{-1}(tan\theta )$

$tan^{-1}\left [ tan\left ( \frac{\pi }{4} -\theta \right ) \right ]=\frac{1}{2}\theta$

$\frac{1}{2}\theta +\theta =\frac{\pi }{4}\;\; \; \; \Rightarrow \; \; \frac{3}{2}\theta =\frac{\pi }{4}$

$\theta =\frac{\pi }{6}\; \; \; \Rightarrow \; \; tan^{-1}x=\frac{\pi }{6}$

$x=tan\left ( \frac{\pi }{6} \right )=\frac{1}{\sqrt{3}}\; \; \; \; \; ...\; \; \: x>0$********************************

NCERT  Example 11 Chapter 2

Show that : $sin^{-1}\left ( \frac{12}{13} \right )+cos^{-1}\left ( \frac{4}{5} \right )+tan^{-1}\left ( \frac{63}{16} \right )=\pi$

$sin^{-1}\left ( \frac{12}{13} \right )=tan^{-1}\left ( \frac{12}{5} \right )$

$cos^{-1}\left ( \frac{4}{5} \right )=tan^{-1}\left ( \frac{3}{4} \right )$
Now given equation becomes

$LHS=tan^{-1}\left ( \frac{12}{5} \right )+tan^{-1}\left ( \frac{3}{4} \right )+tan^{-1}\frac{63}{16}$

$=tan^{-1}\left [ \frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\times \frac{3}{4}} \right ]+tan^{-1}\left ( \frac{63}{16} \right )$

$tan^{-1}\left (- \frac{63}{16} \right )+tan^{-1}\left ( \frac{63}{16} \right )$

$tan^{-1}\left ( \frac{63}{16} \right )+tan^{-1}\left ( -\frac{63}{16} \right )$

$tan^{-1}\left ( \frac{-\frac{63}{16}+\frac{63}{16}}{1+\frac{63}{16}\times \frac{63}{16}} \right )$

$tan^{-1}(0)=\pi$

Reason:  tan 0 = 0 and tan π = 0
$\Rightarrow \; \; \; \pi =tan^{-1}(0)$

Graphs of Trigonometric and Inverse Trigonometric Functions

Graphs of sinx

Graph of and sin-1x

Comparison between sinx and sin-1x

Graphs of cosx and cos-1x

Graphs of tanx and tan-1x]

Graphs of cotx and cot-1x

Graphs of secx and sec-1x

Graphs of cossecx and cosec-1x

THANKS FOR YOUR VISIT