Maths M.C.Q (1 Marks)

4. None of these

Solution:

$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$

$=\tan^{-1}\Bigg(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{2}{11}\times\frac{1}{11}}\Bigg)$

$=\tan^{-1}\Bigg(\frac{\frac{3}{11}}{1-\frac{2}{121}}\Bigg)$

$=\tan^{-1}\Big(\frac{33}{119}\Big)$

1. $\text{x}$

Solution:

Put $\cos^{-1}\text{x}=\text{u}$

$\sin\big[\cot^{-1}\big\{\tan\big(\cos^{-1}\text{x}\big)\big\}\big]$

$=\sin\big[\cot^{-1}\{\tan(\text{u})\}\big]$

$=\sin\Big[\cot^{-1}\Big\{\cot\Big(\frac{\pi}{2}-\text{u}\Big )\Big\}\Big]$

$=\sin\Big[\frac{\pi}{2}-\text{u}\Big]$

$=\cos\text{u}$

$=\text{x}$ $\big(\therefore\ \cos^{-1}\text{x}=\text{u}\Rightarrow\text{x}=\cos\text{u}\big)$

2. $ \frac{\pi}{2}$

Solution:

$\text{x}\in \left ( \frac{3\pi}{2}, 2\pi \right )$

Now, $ \cos^{-1}(\cos \,\text{ x})=2π−\text{x}$

and $ \sin^{-1}(\sin \, \text{x})=\text{x}-2\pi$

$ ∴\cos^{−1}(\cos\text{x})+\sin−1(\sin\text{x})=0$

$\sin−1[\cos{\cos−1(\cos\text{x})+\sin−1(\sin\text{x})}]$

$ =\sin^{−1}{\cos(0)}=\sin^{−1}(1)=\frac{π​}{2}$

3. $\frac{\sqrt{3}}{2}$

2. $\frac{6}{17}$

3. $\tan^{-1}(15)$

4. $\frac{1}{\sqrt{2}}\leq\text{x}\leq1$

3. $\frac{\pi}{4}$

1. $\frac{\pi}{5}$

Solution:

We have, $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{4\pi}{5},$

$\Rightarrow\ \frac{\pi}{2}-\cot^{-1}\text{x}+\frac{\pi}{2}-\cot^{-1}\text{y}=\frac{4\pi}{5}$

$\Rightarrow\ -(\cot^{-1}\text{x}+\cot^{-1}\text{y})=\frac{4\pi}{5}-\pi$

$\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$

$\Rightarrow\ \cot^{-1}\text{x}+\cot^{-1}\text{y}=-\Big(-\frac{\pi}{5}\Big)$

$\Rightarrow\ \cot^{-1}\text{x}+\cot^{-1}\text{y}=\frac{\pi}{5}$

2. $\frac{3\pi}{4}$

2. $-\cos^{-1}\sqrt{1-\text{x}^2}$

Solution:

$\sin^{-1}(\text{x})$

$-\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big),$ for x > 0

Since x takes negative permissible value.

$\sin^{-1}\text{x}=-\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$

2. 1

Solution:

$\tan ^{-1} \frac{\text{x}^4+1-\text{x}^2}{\text{x}^3-\text{x}^5-\text{x}}$

$$$= ^{-1} \frac{3\pi \text{ x}^4+1-\text{x}^2}{\text{x}^3-\text{x}^5-\text{x}}$

$ =−1\text{x}^4 + 1 -\text{x}^2 + \text{x}^3 - \text{x}^5-\text{x} = 0$

= 0 Real roots = 11

3. $\pm\frac{\pi}{6}$

Solution:

We have, $\tan^{-1}(\cot\theta)=2\theta$

$\Rightarrow\tan2\theta=\cot\theta$

$\Rightarrow\frac{2\tan\theta}{1-\tan^2\theta}=\frac{1}{\tan\theta}$

$\Rightarrow2\tan^2\theta=1-\tan^2\theta$

$\Rightarrow3\tan^2\theta=1$

$\Rightarrow\tan^2\theta=\frac{1}{3}$

$\Rightarrow\tan\theta=\pm\frac{1}{\sqrt3}$

1. $\frac{3\pi}{5}$

2. $\frac{1}{5}$

2. $\pi- \cos^{-1}(-\text{x})$

Solution:

Let, $ θ = \cos-1(\text{-x})$

So, $ 0 ≤ θ ≤ π$

$ ⇒ -\text{x} = \cosθ$

$ ⇒ \text{x} = -\cosθ$

$ ⇒ \text{x} = \cos-θ$

Also, $ -π ≤ -θ ≤ 0$

So, $ 0 ≤ π -θ ≤ π$

$ ⇒ -θ = \cos^{-1}(\text{x})$

$ ⇒ θ = \cos^{-1}(\text{x})$

So, $\cos-1(\text{x}) = π – θ$

$ θ = π – \cos-1(\text{x})$

$ ⇒ \cos^{-1}(-\text{x}) = π – \cos^{-1}(\text{x})$

1. $\frac{\pi}{4}$

4. 20⁰

Solution:

sinsin of an angle is positive in first and second quadrants.

$ \Rightarrow \sin ^{ -1 }{ (\sin { { 160 }^{ 0 } } } )$

$\Rightarrow(\sin ^{ -1 }{ (\sin { { (180-20) }^{ 0 } } })=20^0$

2. $\text{e}^{\frac{13\pi}{18}}$

Solution:

Given

$\text{f}(\text{x})=\text{e}^{\cos^{-1}\big\{\sin\big(\text{x}+\frac{\pi}{3}\big)\big\}}$

Then,

$\text{f}\Big(\frac{8\pi}{9}\Big)=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{8\pi}{9}+\frac{\pi}{3}\big)\big\}}$

$=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{11\pi}{9}\big)\big\}}=\text{e}^{\cos^{-1}\big\{\cos\frac{\pi}{2}+\frac{13\pi}{18}\big\}}$ $\Big[\because\ \cos\Big(\frac{\pi}{2}+\theta\Big)=\sin\theta\Big]$

$=\text{e}^{\cos^{-1}\big\{\cos\big(\frac{13\pi}{18}\big)\big\}}$

$=\text{e}^{\frac{13\pi}{18}}$

4. $\frac{2}{\text{x}}$

4. $\frac{\pi}{4}$

2. $ \frac{{ \pi }}{6}$

Solution:

$ \left( {{{\tan }^{ - 1}}\pi + {{\tan }^{ - 1}}\left( {\frac{1}{\pi }} \right)} \right) + {\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$

$=\tan^{-1}\pi+\cot^{-1}\pi+\tan^{-1}\sqrt 3-\sec^{-1}(-2)$

$ [∵\tan^{−1}\frac{1}{\text{y}​}=\cot−1\text{y}]$

$ =\cfrac{\pi}{2}+\tan^{-1}\sqrt 3-\sec^{-1}(-2)\Big[∵\tan^{−1}x+\cot^{−1}\text{x}=\frac{π}{2}\Big]=\frac{π}{2}+\frac{π}{3}-\frac{2π}{3}$
$ =[∵\tan\frac{π}{3}​=\sqrt{3}​;\sec\frac{2π}{3}​=−2]$

$ =\frac{π​}{2}−\frac{π​}{3}$

$ =\frac{\pi}{6}$

None of the given options are correct.

1. $\sqrt{\tan\theta}$

Solution:

Let $\text{y}=\sqrt{\tan\theta}$

Then,

$\Rightarrow\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$

$\Rightarrow\text{u}=\cot^{-1}\text{y}-\tan^{-1}\text{y}$ $\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$

$\Rightarrow2\tan^{-1}\text{y}=\frac{\pi}{2}-\text{u}$

$\Rightarrow\tan^{-1}\text{y}=\frac{\pi}{4}-\frac{\text{u}}{2}$

$\Rightarrow\text{y}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$

$\Rightarrow\sqrt{\tan\theta}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$ $\Big[\because\ \text{y}=\sqrt{\tan\theta}\Big]$

1. $\frac{\pi}{3}$

Solution:

$\theta=\sin^{-1}\{\sin(-600^\circ)\}$

$\theta=\sin^{-1}[\sin(-600^\circ)]$

$\theta=\sin^{-1}[-\sin(180^\circ\times3+60)]$

$\theta=\sin^{-1}[-\{-\sin(60^\circ)\}]$

$\theta=\sin^{-1}(\sin(60^\circ))$

$\theta=\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)$

$\theta=\frac{\pi}{3}$

1. $ \frac { 1 }{ \sqrt { 5 } }$

Solution:

We have, $\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]$

Put $ \cos ^{ -1 }{ \frac { 2 }{ 3 } } =θ$

$ \Rightarrow \cos { \theta } =\frac { 2 }{ 3 }$

$\therefore\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]$

$ \therefore \ \tan\frac{\theta}{2}$

$=\sqrt {\frac {1-\cos {\theta}}{1+\cos{\theta }}}=\sqrt {\frac {1-\frac{2}{3} }{1+{\frac{ 2}{3}}}}​​$

$=\sqrt { \frac { \frac{ 1 }{ 3 } }{ \frac{5}{3} } } =\frac { 1 }{ \sqrt { 5 } }$

2. $\frac{2}{5}$

Solution:

We have, $\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=0$

$\Rightarrow\ \sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}=\cos^{-1}0$

$\Rightarrow\ \sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}=\frac{\pi}{2}$

$\Rightarrow\ \cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\frac{2}{5}$

$\Rightarrow\ \cos^{-1}\text{x}=\cos^{-1}\frac{2}{5}$

$\Big(\because\ \cos^{-1}\text{x}+\sin^{-1}\text{x}=\frac{\pi}{2}\Big)$

$\therefore\ \text{x}=\frac{2}{5}$

4. $ -\cos 4$

Solution:

As for $ \displaystyle \tan ^{ -1 }{ \text{x} }; \text{x}\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right]$

$\cos(\tan^{−1}(\tan4))=\cos(\tan^{−1}(\tan(π−4))$

$ =\cos(π−4)=−\cos4$

4. $ \dfrac{\pi}{2}$

Solution:

Value of $\sin^{-1}(1)\sin\text{x}$ is in vertible form $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$ in this range only $\sin\frac{\pi}{2}=1\sin^{-1}(1)\frac{\pi}{2}.$

2. $\frac{5\pi}{6}$

Solution:

We have, $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)=2\sec^{-1}\sec\frac{\pi}{3}+\sin^{-1}\sin\frac{\pi}{6}$

$=2\frac{\pi}{3}+\frac{\pi}{6}$

$[\because\ \sec^{-1}(\sec\text{x})=\text{x and }\sin^{-1}(\sin\text{x})=\text{x}]$

$=\frac{4\pi+\pi}{6}=\frac{5\pi}{6}$

3. 0.96

Solution:

$\sin\big(2\big(\tan^{-1}0.75\big)\big)=\sin\big(2\tan^{-1}0.75\big)$

$=\sin\Big(\sin^{-1}\frac{2\times0.75}{1+(0.75)^2}\Big)$

$=\sin\big(\sin^{-1}0.96\big)$

$=0 .96$

Hence, the correct answer is option (c).

1. 1

Solution:

Given, $ 1+\text{x}^2+2\text{x}(\sin(\cos^{-1}(\text{y})))=0$

$ 1+\text{x}^2+2\text{x}(\sin(\sin^{-1}(\sqrt{1-\text{y}^2})))=0$

$ 1+\text{x}^2+2\text{x}(\sqrt{1-\text{y}^2})=0$

$ 2\text{x}(\sqrt{1-\text{y}^2})=-(1+\text{x}^2)$

$ 4\text{x}^2(1-\text{y}^2)=1+\text{x}^4+2\text{x}^2$

$ 4\text{x}^2-4\text{x}^2\text{y}^2=1+\text{x}^4+2\text{x}^2$

$ -4\text{x}^{2}(\text{y}^2)=(1-\text{x}^2)^{2}$

Hence solution will be x = 1 and $\text{y}=\frac{1}{2}$

1. $0$

1. π

Solution:

We know, $ \displaystyle \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\frac { \pi }{ 2 }$

$\therefore \sin ^{ -1 }{ \text{x} } +\sin ^{ -1 }{ \frac { 1 }{ \text{x} } } +\cos ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \frac { 1 }{ \text{x} } }$

$ \displaystyle =\frac { \pi }{ 2 } +\frac { \pi }{ 2 } =\pi$

1. $\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$

2. 2

Solution:

$ 2\sin ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x}+1 } } +\cos ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x} } } =\frac { 3\pi }{ 2 }$

​For existence of domain of

$\sin^{-1}\sqrt{\text{x} ^2-\text{x}+1}-1≤\sqrt{\text{x}^2-\text{x}}+1≤1$

$ 0 ≤\text{x}^2-\text{x}+1≤1$

$\text{x}∈[0,1]$

For $\cos^{-1}\sqrt{\text{x}^2-\text{x}0}≤\text{x}^2-\text{x}≤1$

$ ⇒\text{x}^2−\text{x}≥0$

$ ⇒\text{x}−\text{x}≥0$

$\text{x}∈[−∞,0]∪[1,∞]$ Only two points are common in their domains i.e.

0 and 1 which also satisfies the given equation.So option B is correct.

3. $\frac{\pi}{6}$

1. $\sin^2\alpha$

Solution:

$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$

Consider, $\cos^{-1}\frac{\text{x}}{\text{a}}+\cos^{-1}\frac{\text{y}}{\text{b}}=\alpha$

$\Rightarrow\cos^{-1}\bigg(\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}\bigg)$

$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}=\cos\alpha$

$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\cos\alpha=\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}$

Squaring on both sides,

$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\Big(1-\frac{\text{x}^2}{\text{a}^2}\Big)\Big(1-\frac{\text{y}^2}{\text{a}^2}\Big)$

$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{a}^2}+\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}$

$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\cos^2\alpha$

$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\sin^2\alpha$

1. $0$

Solution:

As we know that $\sin{0} = 0\sin0=0\Rightarrow 0 = \sin^{-1}{\left( 0 \right)}$

Hence the value of $ \sin^{-1}{\left( 0 \right)}$ is 0.

3. $ \alpha = 0, \beta = \pi$

Solution:

We know that, $ -\frac{\pi}{2}\le \sin^{-1}\text{x}\le \frac{\pi}{2}$

​Now add $ \frac{\pi}{2}​$

each sides $ \frac{\pi}{2}-\frac{\pi}{2}\le \sin^{-1}\text{x}+\frac{\pi}{2}\le \frac{\pi}{2}+\frac{\pi}{2}$

$ \Rightarrow 0\le \sin^{-1}+\dfrac{\pi}{2}\le \pi$

$ \Rightarrow 0\le \sin^{-1}+(\sin^{-1}\text{x}+\cos^{-1}\text{x})\le \pi$ use the identity

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

$ \therefore 0\le 2\sin^{-1}\text{x}+\cos^{-1}\text{x}\le \pi$

Hence $ \alpha=0, \beta=\pi$

2. $\frac{\pi}{3}$

3. $\frac{\sqrt3}{2}$

Solution:

We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$

$4\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi$

$\Rightarrow4\cos^{-1}\text{x}+\frac{\pi}{2}-\cos^{-1}\text{x}=\pi $

$\Rightarrow3\cos^{-1}\text{x}=\pi-\frac{\pi}{2}$

$\Rightarrow3\cos^{-1}\text{x}=\frac{\pi}{2}$

$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$

$\Rightarrow\text{x}=\cos\frac{\pi}{6}$

4. $\frac{1}{5}$

1. $ \frac {\sqrt {2}}{4}$

Solution:

Given, $ \sin\arcsin { \text{x} } =\frac { \sqrt { 2 } }{ 4 } =\frac { 1 }{ 2\sqrt { 2 } }$

$ \therefore \text{arc}\sin { \text{x} } =\text{arc}\sin \left (\frac {1}{2\sqrt2}\right)=0.36136$

$ \therefore \text{x}=\sin(0.36136)=\frac { 1 }{ 2\sqrt { 2 } }$

1. 0

Solution:

We know,

$\text{tan}^1\text{a}+\text{cot}^{-1}\text{a}=\frac{\pi}{2}$

Therefore,

$\text{cot}(\text{tan}^{−1}\text{a}+\text{cot}^{−1}a)=\text{cot}\frac{\pi}{2}=0$