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

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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
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