MATHS M.C.Q (1 Marks)

3. $\frac{44}{117}$

Solution:

We have:

$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$

$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec }\text{x}-\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$

$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$

subtracting (2) from (1):

$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$

$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$

$\Rightarrow2\cot\text{x}=\frac{117}{22}$

$\Rightarrow\cot\text{x}=\frac{117}{44}$

$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$

$\Rightarrow\tan\text{x}=\frac{44}{117}$

3. $\frac{44}{117}$

Solution:

We have:

$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$$$

$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec}\text{ x}-\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$

$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$

subtracting (2) from (1):

$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$

$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$

$\Rightarrow2\cot\text{x}=\frac{117}{22}$

$\Rightarrow\cot\text{x}=\frac{117}{44}$

$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$

$\Rightarrow\tan\text{x}=\frac{44}{117}$

3. 9.5

Solution:

We have:

$\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^285^\circ+\sin^290^\circ$

$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^2(90^\circ-10^\circ)+\sin^2(90^\circ-5^\circ)+\sin^290^\circ$

$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\cos^210^\circ+\cos^25^\circ+\sin^290^\circ$

$=(\sin^25^\circ+\cos^25^\circ)+(\sin^210^\circ+\cos^210^\circ)+(\sin^215^\circ+\cos^215^\circ)$

$+(\sin^220^\circ+\cos^220^\circ)+(\sin^225^\circ+\cos^225^\circ)+(\sin^230^\circ+\cos^230^\circ)$

$+(\sin^235^\circ+\cos^235^\circ)+(\sin^240^\circ+\cos^240^\circ)+\sin^245^\circ+\sin^290^\circ$

$=1+1+1+1+1+1+1+1+\Big(\frac{1}{\sqrt2}\Big)+(1)^2$ $[\because \sin^2\theta+\cos^2\theta=1]$

$=8+\frac{1}{2}+1$

$=9.5$

2. 0

Solution:

$\cos1^\circ \cos2^\circ \cos3^\circ...\ \cos179^\circ$

$=\cos1^\circ\cos2^\circ\cos3^\circ...\ \cos90^\circ...\ \cos179^\circ$

$=0$ $(\cos90^\circ = 0)$

Hence, the correct answer is option B.

2. $\text{x=y, x}\neq0$

Solution:

We have:

$\sec^2\text{x}=\frac{4\text{xy}}{(\text{x}+\text{y})^2}$

$\Rightarrow\frac{4\text{x}\text{y}}{(\text{x}+\text{y})^2}\geq1 $ $[\therefore\sec^2\text{x}\geq1]$

$\Rightarrow4\text{xy}\geq(\text{x}+\text{y})^2$

$\Rightarrow4\text{xy}\geq\text{x}^2+\text{y}^2+2\text{xy}$

$\Rightarrow2\text{xy}\geq\text{x}^2+\text{y}^2$

$\Rightarrow(\text{x}-\text{y})^2\leq0$

$\Rightarrow(\text{x}-\text{y})\leq0$

$\Rightarrow\text{x}=\text{y}$

For $\text{x}=0,\sec^2\text{x}$ will not be defined,

$\Rightarrow\text{x}\neq 0$

$\therefore\text{x}=\text{y}$

2. $-2\sec\text{x}$

Solution:

$-2\sec\text{x}$

$\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$

$=\sqrt{\frac{(1-\sin\text{x})(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}}+\sqrt{\frac{(1+\sin\text{x})(1+\sin\text{x})}{(1-\sin\text{x})(1+\sin\text{x})}}$

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

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

$=\frac{(1-\sin\text{x})}{-\cos\text{x}}+\frac{(1+\sin\text{x})}{-\cos\text{x}}$ $[\frac{\pi}{2}<\text{x}<\pi,\text{so}\cos\text{x}\text{ will }\text{be }\text{negative}.]$

$-(\sec\text{x}-\tan\text{x})-(\sec\text{x}+\tan\text{x})$

$=-2\sec\text{x}$

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

Solution:

We have:

$\frac{\text{y}+1}{1-\text{y}}=\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$

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

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

$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\frac{\Big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\Big)}{\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)}$ $[\because 0< \text{x }<\frac{\pi}{2}\Rightarrow0< \frac{\pi}{2}<\frac{\pi}{4}, 0 \text{ to } \frac{\pi}{4}\cos\text{x}>\sin\text{x}]$

$\Rightarrow\frac{\frac{\cos\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}} +\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}}{\frac{\cos\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}-\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}}$

$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\frac{1+\tan\frac{\text{x}}{2}}{1-\tan\frac{\text{x}}{2}}$

comparing both the sides: 

$\text{y}=\tan\frac{\text{x}}{2}$

2. 1.

Solution:

We have:

$\sin^6\text{A}+\cos^2\text{A}+3(\sin^2\text{A})(\cos^2\text{A})$

$=(\sin^2\text{A})^3+(\cos^2\text{A})^3+3(\sin^2\text{A})(\cos^2\text{A})\times1$

$=(\sin^2\text{A})^3+(\cos^2\text{A})^3+3(\sin^2\text{A})(\cos^2\text{A})(\sin^2\text{A}+\cos^2\text{A})$

$=(\sin^2\text{A}+\cos^2\text{A})^3$

$=1^3=1$

4. $-1-\cot\text{a}$

Solution:

We have:

$\sqrt{2\cot\alpha+\frac{1}{\sin\alpha}}$

$=\sqrt{\frac{2\cos\alpha}{\sin\alpha}+\frac{1}{\sin^2\alpha}}$

$=\sqrt{\frac{2\sin\alpha\cos\alpha+1}{\sin\alpha}}$

$=\sqrt{\frac{2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha}{\sin\alpha}}$

$=\sqrt{\frac{(\sin\alpha+\cos\alpha)^2}{\sin^2\alpha}}$

$=\sqrt{(1+\cot\alpha)^2}$

$=|1+\cot\alpha|$

$=-(1+\cot\alpha) $ $$ $[\text{ when}\frac{3\pi}{4}<\alpha<\pi,\cot\alpha<-1\Rightarrow\cot\alpha+1<0\big]$

$=-1-\cot\alpha$

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

Solution:

We have:

$\tan\text{x}+\sec\text{x}=\sqrt{3}$ $[0,<\text{x}<\pi]$

$\Rightarrow\sec\text{x}+\tan\text{x}=\sqrt{3}$

$\Rightarrow\frac{1}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}=\sqrt{3}$

$\Rightarrow1+\sin\text{x}=\sqrt{3}\cos\text{x}$

$\Rightarrow(1+\sin\text{x})^2=(\sqrt{3}\cos\text{x})^2$

$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3\cos^2\text{x}$

$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3(1-\sin^2\text{x})$

$\Rightarrow4\sin^2\text{x}+2\sin\text{x}=2$

$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$

$\Rightarrow\sin\text{x}=-1,\frac{1}{2}$

since $0<\text{x}<\pi,\ \sin\text{x}$ cannot be negative.

$\therefore\sin\text{x}=\frac{1}{2}$

$\therefore\text{x}=\frac{\pi}{6}$

2. $\sin1^\circ<\sin1$

Solution:

We know that, 1 radian is approximately $57^\circ$.

Also, the value of $\sin\text{x}$ is always increasing for $0\leq \text{x}\leq 90^\circ$ 

$($or $\sin\text{x}$ is an increasing function for $0\leq \text{x}\leq 90^\circ).$

Now, $1^\circ < 57^\circ$

or $1^\circ< 1 \text{ radian}$

$\therefore\sin 1^\circ < \sin1$

Hence, the correct answer is option B.

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

Solution:

We have:

$\sec\text{x} +\tan\text{x} = \text{k}\cdots(1)$

$\Rightarrow\frac{1}{\sec\text{x} + \tan\text{x}}=\frac{1}{\text{k}}$

$\Rightarrow\frac{\sec^2\text{x}-\tan^2\text{x}}{\sec\text{x}+\tan\text{x}} = \frac{1}{\text{k}}$

$\Rightarrow\frac{(\sec\text{x} + \tan\text{x})(\sec\text{x}-\tan\text{x})}{(\sec\text{x} + \tan\text{x})} = \frac{1}{\text{k}}$

$\therefore\sec\text{x} - \tan\text{x} = \frac{1}{\text{k}}\cdots(2)$

Adding (1) and (2):

$2\sec\text{x}= \text{k} + \frac{1}{\text{k}}$

$\Rightarrow 2\sec\text{x} = \frac{\text{k}^2 + 1}{\text{k}}$

$\Rightarrow \sec\text{x} = \frac{\text{k}^2+1}{2\text{k}}$

$\Rightarrow\frac{1}{\cos\text{x}}= \frac{\text{k}^2 + 1}{2\text{k}}$

$\Rightarrow\cos \text{x} = \frac{2\text{k}}{\text{k}^2 + 1}$

1. $\frac{\sqrt{5}}{\sqrt{6}}$

Solution:

In the fourth quadrant, $\cos\text{x}\text{ and }\sec\text{x}$ are positive.

$\cos\text{x}=\frac{1}{\sec\text{x}}$

 $=\frac{1}{\sqrt{\sec^2\text{x}}}$

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

 $=\frac{1}{\sqrt{1+\Big(-\frac{1}{\sqrt{5}}\Big)^2}}$

$=\frac{1}{\sqrt{\frac{6}{5}}}$

$=\frac{\sqrt{5}}{\sqrt{6}}$

3. $\sec\text{x}=\frac12$

Solution:

1.  $\sin\text{x} = -\frac{1}{5}$ is correct as $-1\leq \sin\text{x} \leq1$

2. $\cos\text{x}=1$ is correct as $-1\leq \cos\text{x}\leq1$

3. $\sec\text{x}=\frac{1}{2}$ is correct as $\text{x}\in [(-\infty, -1)\ \cup (1,\infty)]$

4. $\tan\text{x} = 20$ is correct as $\tan\text{x}$ can take any real value.

Hence, the correct answer is option C.

4. $\text{F}(\text{x})\geq2$

Solution:

$\text{f}(\text{x}) = \cos^2\text{x} + \sec^2\text{x}$

$=\cos^2\text{x} +\sec^2\text{x}-2\cos\text{x}\sec\text{x}+2\cos\text{x}\sec\text{x}$

$=(\sec\text{x}-\cos\text{x})^2 +2$

$\therefore\text{f}\text({x})\geq 2 \ \forall \text{ x }$$\Big[(\sec\text{x}-\cos\text{x})^2\geq 0 \ \forall \text{ x}\Big]$

Hence, the correct option is answer D.

3. 8

Solution:

We have:

$\text{x}\sin45^\circ\cos^260^\circ=\frac{\tan^260^\circ\text{cosec}30^\circ}{\sec45^\circ\cot^230^\circ}$

$\Rightarrow\text{x}\times\Big(\frac{1}{\sqrt2}\Big)\times\Big(\frac{1}{2}\Big)^2=\frac{(\sqrt3)^2\times(2)}{(\sqrt2)\times(\sqrt3)^2}$

$\Rightarrow\frac{\text{x}}{4\sqrt2}=\frac{6}{3\sqrt2}$

$\Rightarrow\text{x}=\frac{6}{3\sqrt2}\times4\sqrt2$

$\Rightarrow\text{x}=8$

2. $2\text{x},\frac{1}{2\text{x}}$

Solution:

We have:

$\sec\text{x} = \text{x} +\frac{1}{4\text{x}}$

$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$

$\Rightarrow1+\tan^2\text{x}$

$=1+\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$

$\Rightarrow\tan^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$

$\Rightarrow\tan^2\text{x}=\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$

$\therefore\tan\text{x}=\pm\Big(\text{x}-\frac{1}{4\text{x}}\Big)$

$\sec\text{x}-\tan\text{x}=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{or}$ 

$=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big[-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\Big]$

$=\frac{1}{2\text{x}}\text{ or } 2\text{x}$

3. 2

Solution:

We have:

$\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{2}+\sin^2\frac{4\pi}{9}$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{8}+\sin^2\frac{7\pi}{18}+\sin^2\frac{8\pi}{2}$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{8}+\sin^2\Big(\frac{7\pi}{18}\Big)+\sin^2\Big(\frac{8\pi}{2}\Big)$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{18}+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{18}\Big)+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{18}\Big)$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{18}+\cos^2\frac{2\pi}{18}+\cos^2\frac{\pi}{18}$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{18}+\cos^2\frac{2\pi}{18}+\cos^2\frac{2\pi}{18}$

$=1+1$

$=2$

1. $-2\text{x},\frac{1}{2\text{x}}$

Solution:

We have:

$\tan\text{x}=\text{x}-\frac{1}{4\text{x}}$

$\Rightarrow\sec^2\text{x}=1+\tan^2\text{x}$

$\Rightarrow\sec^2\text{x}=1+\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$

$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$

$\Rightarrow \sec\text{x}^2 =\Big(\text{x}+\frac{1}{4\text{x}}\Big)^2$

$\therefore \sec\text{x}=\pm\Big(\text{x}+\frac{1}{4\text{x}}\Big)$

$\Rightarrow \sec\text{x}-\tan\text{x} =\Big(\text{x}+\frac{1}{4\text{x}}\Big)- \Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{ or} -\Big(\text{x}+\frac{1}{4\text{x}}\Big)\text{or}- \Big(\text{x}-\frac{1}{4\text{x}}\Big)$

$=\frac{1}{2\text{x}}\text{ or} -2\text{x}$

3. $\tan\text{x}-\sec\text{x}$

Solution:

$\tan\text{x}-\sec\text{x}$

$\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}$

$=\sqrt{\frac{(1-\sin\text{x})(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}}$

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

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

$=\frac{(1-\sin\text{x})}{-\cos\text{x}}$ $[\text{as},\frac{\pi}{2}<\text{x}<\frac{3\pi}{2},\text{so}\cos\theta \text{ will}\text{ be}\text{ negative}]$

$=-(\sec\text{x}-\tan\text{x})$

$=-\sec\text{x} +\tan\text{x}$

2. 1

Solution:

We know that, $\tan(90^\circ-\theta) = \cot\theta$

so,

$\tan89^\circ = \tan(90^\circ-1^\circ) = \cot 1^\circ$

$\tan88^\circ = \tan(90^\circ-2^\circ) = \cot2^\circ$

$\tan87^\circ = \tan(90^\circ-2^\circ) = \cot3^\circ\\\ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .\\\ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .$

$\tan46^\circ = \tan(90^\circ-44^\circ)=\cot44^\circ$

$\therefore\tan1^\circ\tan2^\circ\tan3^\circ...\tan89^\circ$

$=\tan1^\circ\tan2^\circ\tan3^\circ...\ \tan44^\circ\tan45^\circ\tan46^\circ\\...\ \tan87^\circ\tan88^\circ\tan89^\circ$

$= \tan1^\circ\tan2^\circ\tan3^\circ\ ...\ \tan44^\circ\tan45^\circ\tan46^\circ\\...\ \cot3^\circ\cot2^\circ\cot1^\circ$

$ =(\tan1^\circ \cot1^\circ)(\tan2^\circ\cot2^\circ)(\tan3^\circ \cot3^\circ)\\\ ... (\tan44^\circ \cot44^\circ) \tan45^\circ$

$=1$ $(\tan45^\circ =1 \text{ and }\tan\theta\cot\theta = 1)$

Hence, the correct answer is option B.

4. $-\text{cosec}\text{ x} -\cot\text{x}$

Solution:

$-\text{cosec}\text{x} -\cot\text{x}$

$=\sqrt{\frac{1+\cos\text{x}}{1-\cos\text{x}}}$

$=\sqrt{\frac{(1+\cos\text{x})(1+\cos\text{x})}{(1+\cos\text{x})(1+\cos\text{x})}}$

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

$=\sqrt{\frac{(1+\cos\text{x})^2}{\sin^2\text{x}}}$

$=\frac{(1+\cos\text{x})}{-\sin\text{x}}$ $ [\text{as},\pi<\text{x}<2\pi,\text{ so }\sin\text{x} \text{ will}\text{ be}\text{ negative}]$

$=-(\text{cosec}\text{ x}+\cot\text{x})$

$= -\text{cosec}\text{ x } -\cot\text{x}$

2. $\frac{3}{5}$

Solution:

$2\text{cosec}=\frac{1}{2}+2$

$\Rightarrow 2\text{cosec}\text{ x} = \frac{5}{2}$

$\Rightarrow \text{cosec}\text{ x} =\frac{5}{4}$

$\Rightarrow\frac{1}{\sin\text{x}}=\frac{5}{4}$

$\Rightarrow\sin\text{x}=\frac{4}{5}$

Now, $0<\theta<\frac{\pi}{2}$

$\therefore\cos\theta=\sqrt{1-\sin^2\theta}$

$=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$

$=\frac{3}{5}$

2. $\frac{23}{10}$

Solution:

It is given that $\frac{\pi}{2}<\text{A}<\pi$

$3\tan\text{A}+4=0$

$\Rightarrow\tan\text{A}=-\frac{4}{3}$

$\Rightarrow\cot\text{A}=-\frac{3}{4}$

Now, 

$\sec\text{A}=\pm\sqrt{1+\tan^2\text{A}}$

$=\pm\sqrt{1+\frac{16}{9}}$

$=\pm\sqrt{\frac{25}{9}}=\pm\frac{5}{3}$

Also,

$\sin\text{A}=\pm\sqrt{1-\cos^2\text{A}}$

$=\pm\sqrt{1-\frac{9}{25}}$

$=\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}$

$\therefore\sin\text{A}=\frac{4}{5}$ (A lines in 2 nd quadrant)

so,

$2\cot\text{A}-5\cos\text{A}+\sin\text{A}$

$=2\times\Big(-\frac{3}{5}\Big)-5\times\Big(-\frac{3}{5}\Big)+\frac{4}{5}$

$=-\frac{3}{2}+3+\frac{4}{5}$

$=-\frac{15+30+8}{10} $

$=\frac{23}{10}$

Hence,the correct answer is option B.

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

Solution:

We have:

$\tan\text{x}=\frac{1}{\sqrt{7}}$

$\therefore \tan^2\text{x}=\frac{1}{7}$

Now, dividing the numerator and the denominator of $\frac{\text{cosec}^2\text{x}-\sec^2\text{x}}{\text{cosec}^2\text{x}+\sec^2\text{x}{}}$ by $\text{cosec}^2\text{x}=\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}$

$=\frac{1-\frac{1}{7}}{1+\frac{1}{7}}$

$=\frac{6}{8}=\frac{3}{4}$

4. 194

Solution:

We have:

$\tan\text{A}+\cot\text{A}=4$

squaring both the sides:

$(\tan\text{A}+\cot\text{A})^2=4^2$

$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2(\tan\text{A})(\cot\text{A})=16$

$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2=16$

$\Rightarrow\tan^2\text{A}+\cot\text{A}=14$

squaring both the sides again:

$(\tan^2\text{A}+\cot^2\text{A})^2=14^2$

$\tan^4\text{A}+\cot^4\text{A}+2(\tan^2\text{A})(\cot^2\text{A})=196$

$\Rightarrow\tan^4\text{A}+\cot^4\text{A}+2=196$

$\Rightarrow\tan^4\text{A}+\cot^4\text{A}=194$

2. $\frac{2}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$

Solution:

We have:

$\tan\theta + \sec\theta=\text{e}^\text{x}$

$\sec\theta + \tan\theta = \text{e}^\text{x}\cdots(1)$

$\Rightarrow\frac{1}{\sec\theta+\tan\theta}=\frac{1}{\text{e}^\text{x}}$

$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta + \tan\theta}=\frac{1}{\text{e}^\text{x}}$

$\Rightarrow\frac{(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(\sec\theta + \tan\theta)}=\frac{1}{\text{e}^\text{x}}$

$\therefore \sec\theta - \tan\theta=\frac{1}{\text{e}^\text{x}}\cdots(2)$

Adding (1) and (2): 

$2\sec\theta = \text{e}^\text{x}+ \frac{1}{\text{e}^\text{x}}$

$\Rightarrow 2\sec\theta = \frac{(\text{e}^\text{x})^2+1}{\text{e}^\text{x}}$

$\Rightarrow \sec\theta = \frac{\text{e}^{2\text{x}}+1}{2\text{e}^\text{x}}$

$\Rightarrow\sec\theta=\frac{1}{2}\times\frac{\text{e}^{2\text{x}}+1}{2\text{e}^\text{x}}$

$\Rightarrow\sec\theta = \frac{1}{2}\times(\text{e}^\text{x}+\text{e}^\text{-x})$

$\Rightarrow\frac{1}{\cos\theta}=\frac{\text{e}^\text{x}+\text{e}^\text{x}}{2}$

$\Rightarrow\cos\theta= \frac{2}{\text{e}^\text{x}+\text{e}^\text{-x}}$

1. $\theta,\phi$

Solution:

We have:

$\text{x}=\text{r}\sin\theta\cos\phi,\text{y}=\text{r}\sin\theta\sin\phi\text{ and }\text{z}=\text{r}\cos\theta,$

$\therefore\text{x}^2+\text{y}^2+\text{z}^2$

$=(\text{r}\sin\theta\cos\phi)^2+(\text{r}\sin\theta\sin\phi)^2+(\text{r}\cos\theta)^2$

$=\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi+\text{r}^2\cos^2\theta$

$=\text{r}^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+\text{r}^2\cos^2\theta$

$=\text{r}^2\sin^2\theta\times1+\text{r}^2\cos^2\theta$

$=\text{r}^2\sin^2\theta+\text{r}\cos^2\theta$

$=\text{r}^2(\sin^2\theta+\cos^2\theta)$

$=\text{r}^2\times1$

$=\text{r}^2$

Thus, $\text{x}^2+\text{y}^2+\text{z}^2$ is independent of $\theta\text{ and }\phi$