University Maths Solution

Maths Question
Question 1

$\text{Establish }(\tan \theta +\cot \theta )\tan \theta ={{\sec }^{2}}\theta$

Question 2

$\text{Establish }\frac{\sin t}{1-\cos t}=\csc t+\cot t$

Question 3

$\text{Establish }\cot \alpha (\sec \alpha +\cos ec\alpha )$

Question 4

$\text{Establish }\frac{1}{\sin \alpha +\cos \alpha }=\frac{\tan \alpha +\cot \alpha }{\sec \alpha +\operatorname{cosec}\alpha }$

Question 5

$\text{Establish the identity }\frac{1-\cos \alpha }{\sin \alpha }=\frac{1}{\operatorname{cosec}\alpha +\cot \alpha }$

Question 6

$\text{Establish that }\frac{{{\tan }^{2}}\theta +{{\cos }^{2}}\theta }{\sec \theta +\sin \theta }=\sec \theta -\sin \theta$

Question 7

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

Question 8

$\text{Establish that }\frac{1-\sin \theta +\cos \theta }{1-\sin \theta }=\frac{1+\sin \theta +\cos \theta }{\cos \theta }$

Question 9

$\text{Establish that }\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\operatorname{cosec}\theta$

Question 10

$\text{Establish that }\frac{\tan u-\tan v}{1+\tan u\tan v}=\frac{\cot v-\cot u}{1+\cot u\cot v}$

Question 11

$\text{Establish that }\frac{1}{\tan \beta +\cot \beta }=\sin \beta \cos \beta$

Question 12

$\text{Find the values in terms of }p\text{ and }q\text{ of }\frac{p\cos \theta +q\sin \theta }{p\cos \theta -q\sin \theta }\text{ where }\cot \theta =\frac{p}{q}$

Question 13

$\text{If }\cos \theta =a,\text{ find the value of }\operatorname{cosec}\left( \tfrac{\pi }{2}+\theta \right)\text{ and }\sin \left( \tfrac{3\pi }{2}-\theta \right)$

Question 14

\begin{align} & \text{if }\cos \theta =\tfrac{4}{5}\text{ (i) }\sin \theta \text{ (ii) }\operatorname{cosec}\theta \text{ (iii) }\theta \\ & \text{If }\sin \theta =-\tfrac{12}{13}\text{ (i) cos}\theta \text{ (ii) }\tan \theta \\\end{align}

Question 15

\begin{align} & \text{Write down the general solution of the trigonometric equation} \\ & \sin (\theta +{{45}^{\circ }})=\frac{1}{2} \\\end{align}

Question 16

\begin{align} & \text{Write down the general solution of the trigonometric equations} \\ & \cos (2\theta -{{30}^{\circ }})=0 \\\end{align}

Question 17

\begin{align} & \text{Write down the general solution of the trigonometric equations} \\ & \tan 3\theta =\sqrt{3} \\\end{align}

Question 18

\begin{align} & \text{Write down the general solution of the trigonometric equations} \\ & 2\sin 2\theta +1=0 \\\end{align}

Question 19

\begin{align} & \text{Write down the general solution of the trigonometric equations} \\ & 2\cos ({{90}^{\circ }}-\theta )=\sqrt{2} \\\end{align}

Question 20

$\text{Verify }\cos \left( \theta +\frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}(\cos \theta -\sin \theta )$

Question 21

$\text{Verify }\operatorname{cosec}\theta \left( \frac{\pi }{4}-u \right)=\sec u$

Question 22

$\text{Verify }\cot \left( t-\frac{\pi }{3} \right)=\frac{\sqrt{3}(\tan t+3)}{\tan t-\sqrt{3}}$

Question 23

$\text{Verify }\sec 2m=\frac{{{\sec }^{2}}m}{2-{{\sec }^{2}}m}$

Question 24

\begin{align} & \text{Obtain the first general solution of the equation and then use the general } \\ & \text{solution to find all solution in the range }{{0}^{\circ }}\le \theta \le {{360}^{\circ }} \\ & 8{{\sin }^{2}}\theta -6\cos \theta =3 \\\end{align}

Question 25

\begin{align} & \text{Obtain the first general solution of the equation and then use the general } \\ & \text{solution to find all solution in the range }{{0}^{\circ }}\le \theta \le {{360}^{\circ }} \\ & 7{{\sec }^{2}}\theta =6\tan \theta +8 \\\end{align}

Question 26

\begin{align} & \text{Obtain the first general solution of the equation and then use the general } \\ & \text{solution to find all solution in the range }{{0}^{\circ }}\le \theta \le {{360}^{\circ }} \\ & 3\sin 3\theta -\operatorname{cosec}3\theta +2=0 \\\end{align}

Question 27

\begin{align} & \text{Obtain the first general solution of the equation and then use the general } \\ & \text{solution to find all solution in the range }{{0}^{\circ }}\le \theta \le {{360}^{\circ }} \\ & 2{{\cos }^{2}}\theta =\sin \theta +1 \\\end{align}