ncert-class12-day14-chap2-inverse-trigonometric-functions-ncert-exemplar-

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<marquee behavior="scroll" direction="left">Acknowledgement and Disclaimer</marquee></p>

<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

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### <Primary style = "float:center; text-align:center;"><B>Inverse Trigonometric Functions: <br> Exemplar Problems</B></Primary>

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Question</B></Primary>

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Prove that $\tan \left(\cot ^{-1} x\right)=\cot \left(\tan ^{-1} x\right)$. State with reason whether the equality is valid for all values of $x$.

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सिद्ध कीजिए कि $\tan \left(\cot ^{-1} x\right)=\cot \left(\tan ^{-1} x\right)$. कारण सहित बताइए कि क्या यह $x$ के सभी मानों के लिए सत्य है।

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Answer</B></Primary>

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**Answer**

Let $\cot ^{-1} x=\theta$. Then $\cot \theta=x$

or, $\tan \left(\frac{\pi}{2}-\theta\right)=x \Rightarrow \tan ^{-1} x=\frac{\pi}{2}-\theta$

So $\tan \left(\cot ^{-1} x\right)=\tan \theta=\cot \left(\frac{\pi}{2}-\theta\right)=\cot \left(\frac{\pi}{2}-\cot ^{-1} x\right)=\cot \left(\tan ^{-1} x\right)$

The equality is valid for all values of $x$ since $\tan ^{-1} x$ and $\cot ^{-1} x$ are true for $x \in \mathbf{R}$.

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**हल**

मान लीजिए $\cot ^{-1} x=\theta$. तब $\cot \theta=x$

या, $\tan \frac{\pi}{2}-\theta=x \Rightarrow \tan ^{-1} x=\frac{\pi}{2}-\theta$ या $\tan \left(\cot ^{-1} x\right)=\tan \left(\frac{\pi}{2}-\tan ^{-1} x\right)=\cot \left(\tan ^{-1} x\right)$

इसलिए $\tan \left(\cot ^{-1} x\right)=\tan \theta=\cot \left(\frac{\pi}{2}-\theta\right)=\cot \left(\frac{\pi}{2}-\cot ^{-1} x\right)=\cot \left(\tan ^{-1} x\right)$
यह समता $x$ के सभी मानों के लिए सत्य है क्योंकि $x \in \mathbf{R}$ के लिए $\tan ^{-1} x$ तथा $\cot ^{-1} x$ सत्य है।

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Question</B></Primary>

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Find the value of $\sec \left(\tan ^{-1} \frac{y}{2}\right)$.

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$ \sec \left(\tan ^{-1} \frac{y}{2}\right)$ का मान ज्ञात कीजिए

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

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**Answer**

Solution Let $\tan ^{-1} \frac{y}{2}=\theta$, where $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. So, $\tan \theta=\frac{y}{2}$,
which gives $\sec \theta=\frac{\sqrt{4+y^2}}{2}$.
Therefore, $\sec \left(\tan ^{-1} \frac{y}{2}\right)=\sec \theta=\frac{\sqrt{4+y^2}}{2}$.

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**हल**

मान लीजिए $\tan ^{-1} \frac{y}{2}=\theta$, जहाँ $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. इसलिए , $\tan \theta=\frac{y}{2}$,
जिससे $\sec \theta=\frac{\sqrt{4+y^2}}{2}$ प्राप्त होता है।
इसलिए, $\sec \left(\tan ^{-1} \frac{y}{2}\right)=\sec \theta=\frac{\sqrt{4+y^2}}{2}$.

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

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Find value of $\tan \left(\cos ^{-1} x\right)$ and hence evaluate $\tan \left(\cos ^{-1} \frac{8}{17}\right)$.

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$ \tan \left(\cos ^{-1} x\right)$ का मान ज्ञात कीजिए और फिर $\tan \cos ^{-1} \frac{8}{17}$ परिकलित कीजिए।

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**Answer**

Let $\cos ^{-1} x=\theta$, then $\cos \theta=x$, where $\theta \in[0, \pi]$

Therefore, $\quad \tan \left(\cos ^{-1} x\right)=\tan \theta=\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}=\frac{\sqrt{1-x^2}}{x}$.
Hence $\tan \left(\cos ^{-1} \frac{8}{17}\right)=\frac{\sqrt{1-\left(\frac{8}{17}\right)^2}}{\frac{8}{17}}=\frac{15}{8}$.

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**हल**

मान लीजिए $\cos ^{-1} x=\theta$, तब $\cos \theta=x$, जहाँ $\theta \in[0, \pi]$

इसलिए $\quad \tan \left(\cos ^{-1} x\right)=\tan \theta=\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}=\frac{\sqrt{1-x^2}}{x}$

अतः $\tan \left(\cos ^{-1} \frac{8}{17}\right)=\frac{\sqrt{1-\left(\frac{8}{17}\right)^2}}{\frac{8}{17}}=\frac{15}{8}$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Question</B></Primary>

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Find the value of $\sin \left[2 \cot ^{-1}\left(\frac{-5}{12}\right)\right]$

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$ \sin 2 \cot ^{-1} \frac{-5}{12}$ का मान ज्ञात कीजिए

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**Answer**

Let $\cot ^{-1}\left(\frac{-5}{12}\right)=y$.

Then $\cot y=\frac{-5}{12}$.

Now $\sin \left[2 \cot ^{-1}\left(\frac{-5}{12}\right)\right]=\sin 2 y$

$
=2 \sin y \cos y=2\left(\frac{12}{13}\right)\left(\frac{-5}{13}\right) \quad\left[\text { since cot } y<0 \text {, so } y \in\left(\frac{\pi}{2}, \pi\right)\right]
$

$
=\frac{-120}{169}
$

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**हल**

मान लीजिए $\cot ^{-1}\left(\frac{-5}{12}\right)=y$. तब $\cot y=\frac{-5}{12}$

अब $\sin 2 \cot ^{-1} \frac{-5}{12}=\sin 2 y$

$
=2 \sin y \cos y=2 \frac{12}{13} \frac{-5}{13} \quad\left[\text { क्योंकि } \cot y<0 \text {, so } y \in\left(\frac{\pi}{2}, \pi\right)\right]  =\frac{-120}{169}
$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Question</B></Primary>

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Evaluate $\cos \left[\sin ^{-1} \frac{1}{4}+\sec ^{-1} \frac{4}{3}\right]$

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$ \cos \left[\sin ^{-1} \frac{1}{4}+\sec ^{-1} \frac{4}{3}\right]$ का मान ज्ञात कीजिए

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**Answer**

$\cos \left[\sin ^{-1} \frac{1}{4}+\sec ^{-1} \frac{4}{3}\right]=\cos \left[\sin ^{-1} \frac{1}{4}+\cos ^{-1} \frac{3}{4}\right]$

$
\begin{aligned}
& =\cos \left(\sin ^{-1} \frac{1}{4}\right) \cos \left(\cos ^{-1} \frac{3}{4}\right)-\sin \left(\sin ^{-1} \frac{1}{4}\right) \sin \left(\cos ^{-1} \frac{3}{4}\right) \\\\
& =\frac{3}{4} \sqrt{1-\left(\frac{1}{4}\right)^2}-\frac{1}{4} \sqrt{1-\left(\frac{3}{4}\right)^2} \\\\
& =\frac{3}{4} \frac{\sqrt{15}}{4}-\frac{1}{4} \frac{\sqrt{7}}{4}=\frac{3 \sqrt{15}-\sqrt{7}}{16} .
\end{aligned}
$
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Find the values of $x$ which satisfy the equation

$$
\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x .
$$

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$ x$ के वे मान ज्ञात कीजिए जो समीकरण $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$ को संतुष्ट करते हैं।

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**Answer**

$$
\begin{aligned}
& \sin \left(\sin ^{-1} x+\sin ^{-1}(1-x)\right)=\sin \left(\cos ^{-1} x\right) \\\\
& \Rightarrow \sin \left(\sin ^{-1} x\right) \cos \left(\sin ^{-1}(1-x)\right)+\cos \left(\sin ^{-1} x\right) \sin \left(\sin ^{-1}(1-x)\right)=\sin \left(\cos ^{-1} x\right) \\\\
& \Rightarrow x \sqrt{1-(1-x)^2}+(1-x) \sqrt{1-x^2}=\sqrt{1-x^2} \\\\
& \Rightarrow x \sqrt{2 x-x^2}+\sqrt{1-x^2}(1-x-1)=0 \\\\
& \Rightarrow x\left(\sqrt{2 x-x^2}-\sqrt{1-x^2}\right)=0 \\\\
& \Rightarrow x=0 \quad \text { or } \quad 2 x-x^2=1-x^2 \\\\
& \Rightarrow x=0 \quad \text { or } \quad x=\frac{1}{2} .
\end{aligned}
$$

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Solve the equation $\sin ^{-1} 6 x+\sin ^{-1} 6 \sqrt{3} x=-\frac{\pi}{2}$

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समीकरण $\sin ^{-1} 6 x+\sin ^{-1} 6 \sqrt{3} x=-\frac{\pi}{2}$ को हल कीजिए।

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**Answer**

From the given equation, we have $\sin ^{-1} 6 x=-\frac{\pi}{2}-\sin ^{-1} 6 \sqrt{3} x$

$$
\begin{array}{ll}
\Rightarrow & \sin \left(\sin ^{-1} 6 x\right)=\sin \left(-\frac{\pi}{2}-\sin ^{-1} 6 \sqrt{3} x\right) \\\\
\Rightarrow & 6 x=-\cos \left(\sin ^{-1} 6 \sqrt{3} x\right) \\\\
\Rightarrow \quad & 6 x=-\sqrt{1-108 x^2} . \text { Squaring, we get } \\\\
& 36 x^2=1-108 x^2 \\\\
\Rightarrow & 144 x^2=1 \quad \Rightarrow x= \pm \frac{1}{12}
\end{array}
$$

Note that $x=-\frac{1}{12}$ is the only root of the equation as $x=\frac{1}{12}$ does not satisfy it.

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हल दिए गए समीकरण को $\sin ^{-1} 6 x=-\frac{\pi}{2}-\sin ^{-1} 6 \sqrt{3} x$ के रूप में लिख सकते हैं।
$$
\begin{array}{ll}
\Rightarrow & \sin \left(\sin ^{-1} 6 x\right)=\sin \left(-\frac{\pi}{2}-\sin ^{-1} 6 \sqrt{3} x\right) \\\\
\Rightarrow & 6 x=-\cos \left(\sin ^{-1} 6 \sqrt{3} x\right) \\\\
\Rightarrow & 6 x=-\sqrt{1-108 x^2}
\end{array}
$$

वर्ग करने पर प्राप्त होता है $36 x^2=1-108 x^2$
$$
\Rightarrow \quad 144 x^2=1 \quad \Rightarrow x= \pm \frac{1}{12}
$$

ध्यान दीजिए कि केवल $x=-\frac{1}{12}$ ही समीकरण का हल है क्योंकि $x=\frac{1}{12}$ इसे संतुष्ट नहीं करता है।

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Show that

$$
2 \tan ^{-1}\left\\{\tan \frac{\alpha}{2} \cdot \tan \left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right\\}=\tan ^{-1} \frac{\sin \alpha \cos \beta}{\cos \alpha+\sin \beta}
$$
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**Answer**

L.H.S. $=\tan ^{-1} \frac{2 \tan \frac{\alpha}{2} \cdot \tan \left(\frac{\pi}{4}-\frac{\beta}{2}\right)}{1-\tan ^2 \frac{\alpha}{2} \tan ^2\left(\frac{\pi}{4}-\frac{\beta}{2}\right)} \quad\left(\right.$ since $\left.2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right)$

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

$
=\tan ^{-1} \frac{2 \tan \frac{\alpha}{2} \cdot\left(1-\tan ^2 \frac{\beta}{2}\right)}{\left(1+\tan \frac{\beta}{2}\right)^2-\tan ^2 \frac{\alpha}{2}\left(1-\tan \frac{\beta}{2}\right)^2}
$
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**Answer**

$\begin{aligned} & =\tan ^{-1} \frac{2 \tan \frac{\alpha}{2}\left(1-\tan ^2 \frac{\beta}{2}\right)}{\left(1+\tan ^2 \frac{\beta}{2}\right)\left(1-\tan ^2 \frac{\alpha}{2}\right)+2 \tan \frac{\beta}{2}\left(1+\tan ^2 \frac{\alpha}{2}\right)} \\\\ & =\tan ^{-1} \frac{\frac{2 \tan \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{2}} \frac{1-\tan ^2 \frac{\beta}{2}}{1+\tan ^2 \frac{\beta}{2}}}{\frac{1-\tan ^2 \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{2}}+\frac{2 \tan \frac{\beta}{2}}{1+\tan ^2 \frac{\beta}{2}}} \\\\ & =\tan ^{-1}\left(\frac{\sin \alpha \cos \beta}{\cos \alpha+\sin \beta}\right)=\text { R.H.S. } \\\\ & \end{aligned}$
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The value of the expression $\sin \left[\cot ^{-1}\left(\cos \left(\tan ^{-1} 1\right)\right)\right]$ is

(A) 0

(B) 1

(D) $\sqrt{\frac{2}{3}}$.

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व्यंजक $\sin \left[\cot ^{-1}\left(\cos \left(\tan ^{-1} 1\right)\right)\right]$ का मान है

(A) 0

(B) 1

(D) $\sqrt{\frac{2}{3}}$

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**Answer**

(D) is the correct answer.

$
\sin \left[\cot ^{-1}\left(\cos \frac{\pi}{4}\right)\right]=\sin \left[\cot ^{-1} \frac{1}{\sqrt{2}}\right]=\sin \left[\sin ^{-1} \sqrt{\frac{2}{3}}\right]=\sqrt{\frac{2}{3}}
$

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**हल**

सही उत्तर (D) है। क्योंकि

$
\sin \left[\cot ^{-1}\left(\cos \frac{\pi}{4}\right)\right]=\sin \left[\cot ^{-1} \frac{1}{\sqrt{2}}\right]=\sin \left[\sin ^{-1} \sqrt{\frac{2}{3}}\right]=\sqrt{\frac{2}{3}}
$

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The equation $\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ has

(A) no solution

(B) unique solution

(D) two solutions

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समीकरण $\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(A) का काई हल नहीं है

(B) का केवल एक मात्र हल है

(D) के दो हल हैं

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**Answer**

(B) is the correct answer. We have

$
\tan ^{-1} x-\cot ^{-1} x=\frac{\pi}{6} \text { and } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}
$

Adding them, we get $2 \tan ^{-1} x=\frac{2 \pi}{3}$

$
\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \text { i.e., } x=\sqrt{3} \text {. }
$
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**हल**

सही उत्तर (B) है। क्योंकि
$\tan ^{-1} x-\cot ^{-1} x=\frac{\pi}{6}$ तथा $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$.

इनको जोड़ने पर हमें $2 \tan ^{-1} x=\frac{2 \pi}{3}$ प्राप्त होता है
इसलिए $\quad \Rightarrow \tan ^{-1} x=\frac{\pi}{3}$ अर्थात् $x=\sqrt{3}$.

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If $\alpha \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \beta$, then

(A) $\alpha=\frac{-\pi}{2}, \beta=\frac{\pi}{2}$

(B) $\alpha=0, \beta=\pi$

(D) $\alpha=0, \beta=2 \pi$
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40 यदि $\alpha \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \beta$, तब

(A) $\alpha=\frac{-\pi}{2}, \beta=\frac{\pi}{2}$

(B) $\alpha=0, \beta=\pi$

(D) $\alpha=0, \beta=2 \pi$

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**Answer**

(B) is the correct answer. We have $\frac{-\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$

$$
\begin{array}{ll}
\Rightarrow & \frac{-\pi}{2}+\frac{\pi}{2} \leq \sin ^{-1} x+\frac{\pi}{2} \leq \frac{\pi}{2}+\frac{\pi}{2} \\\\
\Rightarrow & 0 \leq \sin ^{-1} x+\left(\sin ^{-1} x+\cos ^{-1} x\right) \leq \pi \\\\
\Rightarrow & 0 \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \pi
\end{array}
$$

</div>
<div  style='float: right; width:49%; text-align: left;font-size:21px'>

**हल**

सही उत्तर (B) है। दिया गया है कि $\frac{-\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$

$$
\begin{aligned}
& \Rightarrow \quad \frac{-\pi}{2}+\frac{\pi}{2} \leq \sin ^{-1} x+\frac{\pi}{2} \leq \frac{\pi}{2}+\frac{\pi}{2} \\\\
& \Rightarrow \quad 0 \leq \sin ^{-1} x+\left(\sin ^{-1} x+\cos ^{-1} x\right) \leq \pi \\\\
& \Rightarrow \quad 0 \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \pi
\end{aligned}
$$

</div>

</main>

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Question</B></Primary>

<hr>

<main>

The value of $\tan ^2\left(\sec ^{-1} 2\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right)$ is

(A) 5

(B) 11

(D) 15

</div>
<div  style='float: right; width:49%; text-align: left;font-size:21px'>

$41 \tan ^2\left(\sec ^{-1} 2\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right)$ का मान है

(A) 5

(B) 11

(D) 15

</div>

</main>

======

<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Answer</B></Primary>

<hr>

<main>

**Answer**

(B) is the correct answer.

$ \tan ^2\left(\sec ^{-1} 2\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right)$

$=\sec ^2\left(\sec ^{-1} 2\right)-1+\operatorname{cosec}^2\left(\operatorname{cosec}^{-1} 3\right)-1 $

$ =2^2 \times 1+3^2-2=11 .$
</div>
<div  style='float: right; width:49%; text-align: left;font-size:21px'>

**हल**

सही उत्तर $(\mathrm{B})$ है।

$ \tan ^2\left(\sec ^{-1} 2\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right)$

$=\sec ^2\left(\sec ^{-1} 2\right)-1+\operatorname{cosec}^2\left(\operatorname{cosec}^{-1} 3\right)-1 $

$ =2^2 \times 1+3^2-2=11 .$

</div>

</main>

======

<Success style = "float:center; text-align:center; font-size: 22px;"><B>ITF: Exemplar Problems</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Answer</B></Primary>

<hr>

<main>

**Answer**

</div>
<div  style='float: right; width:49%; text-align: left;font-size:21px'>

Hindi

</div>

</main>

======
#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Don't forget to:</B></Primary>

- Register yourself and give subject-wise mock test.

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- Register yourself and ask any doubt regarding these video lectures or any subject doubt.

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- Here you can find "how to videos"

👉 https://www.youtube.com/watch?v=MT3b32wKnmU

</div>

- अपना रजिस्ट्रेशन करें और विषयवार मॉक टेस्ट दें।

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- यहां आप "पंजीकरण कैसे करें" पा सकते हैं -

👉 https://www.youtube.com/watch?v=MT3b32wKnmU

</div>

======

$\quad$

### <Primary>Thank you</Primary>