TRIGONOMETRY

TRIGONOMETRY

11.1 TRIGONOMETRY :

Trigonometry means, the science which deals with the measurement of triangles.

11.1 (a) Trigonometric Ratios :

A right angled triangle is shown in Figure. $\mathrm{\angle }B$ Is of ${90}^{\circ }$ Side opposite to $\mathrm{\angle }B$ be called hypotenuse. There are two other angles i.e. $\mathrm{\angle }A$ and $\mathrm{\angle }C$. It we consider $\mathrm{\angle }C$ as $\theta$, then opposite side to this angle is called perpendicular and side adjacent to $\theta$ is called base.

TRIGONOMETRY

(i) Six Trigonometry Ratio are :

$\begin{array}{cc}\sin \theta =\frac{\text{ Perpenicular }}{\text{ Hypotenuse }}=\frac{P}{H}=\frac{AB}{AC}& cosec\theta =\frac{\text{ Hypoteuse }}{\text{ Perpendicular }}=\frac{H}{P}=\frac{AC}{AB}\\ \cos \theta =\frac{\text{ Base }}{\text{ Hypotenuse }}=\frac{B}{H}=\frac{BC}{AC}& \sec \theta =\frac{\text{ Hypotenuse }}{\text{ Base }}=\frac{H}{B}=\frac{AC}{BC}\\ \tan \theta =\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{P}{B}=\frac{AB}{BC}& \cot \theta =\frac{\text{ Base }}{\text{ Parpendicular }}=\frac{B}{P}=\frac{BC}{AB}\end{array}$

TRIGONOMETRY

(ii) Interrelationship is Basic Trigonometric Ratio :

$\begin{array}{cc}\tan \theta =\frac{1}{\cot \theta }\Rightarrow & \cot \theta =\frac{1}{\tan \theta }\\ \cos \theta =\frac{1}{\sec \theta }\Rightarrow & \sec \theta =\frac{1}{\cos \theta }\\ \sin \theta =\frac{1}{cosec\theta }\Rightarrow & cosec\theta =\frac{1}{\sin \theta }\end{array}$

We also observe that

$\tan \theta =\frac{\sin \theta }{\cos \theta }\Rightarrow \cot \theta =\frac{\cos \theta }{\sin \theta }$

TRIGONOMETRY

11.1 (b) Trigonometric Table :

TRIGONOMETRY

11.1 (c) Trigonometric Identities :

(i) ${\sin }^2\theta +{\cos }^2\theta =1$

(A) ${\sin }^2\theta =1-{\cos }^2\theta$

(B) ${\cos }^2\theta =1-\sin 2\theta$

(ii) $1+{\tan }^2\theta ={\sec }^2\theta$

(A) ${\sec }^2\theta -1={\tan }^2\theta$

(B) ${\sec }^2\theta -{\tan }^2\theta =1$

(C) ${\tan }^2\theta -{\sec }^2\theta =-1$

(iii) $1+{\cot }^2\theta =cose{c}^2\theta$

(A) $cose{c}^2\theta -1={\cot }^2\theta$

(B) $cose{c}^2\theta -{\cot }^2\theta =1$

(C) ${\cot }^2\theta -cose{c}^2\theta =-1$

TRIGONOMETRY

11.1 (d) Trigonometric Ratio of Complementary Angles :

TRIGONOMETRY

ILLUSTRATIONS :

Ex. 1 In the given triangle $AB=3cm$ and $AC=5cm$. Find all trigonometric ratios.

TRIGONOMETRY

Sol. Using Pythagoras theorem

$\begin{array}{rl}& A{C}^2=A{B}^2+B{C}^2\\ & \Rightarrow 5^2=3^2+p^2\\ & \Rightarrow 16=p^2\Rightarrow P=cm\\ & \text{ Here }P=4cm,B=3cm,H=5cm\\ & \therefore \sin \theta =\frac{P}{H}=\frac{4}{5}\\ & \cos \theta =\frac{B}{H}=\frac{3}{5}\\ & \tan \theta =\frac{P}{B}=\frac{4}{3}\end{array}$

TRIGONOMETRY

$\begin{array}{rl}& \cot \theta =\frac{B}{P}=\frac{3}{4}\\ & \sec \theta =\frac{H}{B}=\frac{5}{3}\\ & cosec\theta =\frac{H}{P}=\frac{5}{4}\end{array}$

TRIGONOMETRY

Ex. 2 If $\tan \theta =\frac{m}{n}$, then find $\sin \theta$.

TRIGONOMETRY

Sol. Let $P=m\alpha$ and $B=n\alpha$

$\begin{array}{cc}\therefore & \tan \theta =\frac{P}{B}=\frac{m}{n}\\ & H^2=P^2+B^2\\ & H^2=m^2{\alpha }^2+n^2{\alpha }^2\\ & H=\alpha \sqrt{m^2+n^2}\\ \therefore & \tan \theta =\frac{P}{H}=\frac{ma}{a\sqrt{m^2+n^2}}\\ & \sin \theta =\frac{m}{\sqrt{m^2+n^2}}\end{array}$

TRIGONOMETRY

Ex. 3 If $cosecA=\frac{13}{5}$ the prove than ${\tan }^{2}A-sin{g}^{2}A={\sin }^{4}A{\sec }^{2}A$.

TRIGONOMETRY

Sol. We hare coses $A=\frac{13}{5}=\frac{\text{ Hypotenuse }}{\text{ Perpendicular }}$

So, we draw a right triangle $ABC$, right angled at $C$ such that hypotenuse $AB=13$ units and perpendicular

$BC=5$ units

$B$ Pythagoras theorem,

$\begin{array}{rl}& A{B}^2=B{C}^2+A{C}^2\Rightarrow (13{)}^2=(5{)}^2+A{C}^2\\ & A{C}^2=169-25=144\\ & AC=\sqrt{144}=12\text{ units }\\ & \tan A=\frac{BC}{AC}=\frac{5}{12}\\ & \sin A=\frac{BC}{AB}=\frac{5}{13}\\ & \text{ and }\sec A=\frac{AB}{AC}=\frac{13}{12}\end{array}$

TRIGONOMETRY

L.H.S. ${\tan }^{2}A-Si{n}^{2}A$

$=(\frac{5}{12}{)}^2-(\frac{5}{13}{)}^2$

$=\frac{25}{144}-\frac{25}{169}$

$=\frac{25(169-144)}{144\times 169}$

$=\frac{25\times 25}{144\times 169}$

TRIGONOMETRY

R.H.S. $={\sin }^{4}A\times {\sec }^{2}A$

$=(\frac{5}{13}{)}^4\times (\frac{13}{12}{)}^2$

$=\frac{5^4\times {13}^2}{{13}^4\times {12}^2}$

$=\frac{5^4}{{13}^2\times {12}^2}$

$=\frac{25\times 25}{144\times 169}$

So, L.H.S. = R.H.S.

Hence Proved.

TRIGONOMETRY

Ex. 4 In $\mathrm{△}ABC$, right angled at $B.AC+AB=9cm$. Determine the value of $\cot C,cosecC,\sec C$.

TRIGONOMETRY

Sol. In $\mathrm{△}ABC$, we have $(AC{)}^2=(AB{)}^2+B{C}^2$

$\Rightarrow (9-AB{)}^2=A{B}^2+(3{)}^2$ $[\because AC+AB=9cm\Rightarrow AC=9-AB]$

$\Rightarrow (81+A{B}^2-18AB=A{B}^2+9.$

$\Rightarrow 72-18AB=0$

$\Rightarrow AB=\frac{72}{18}=4cm$.

TRIGONOMETRY

Now, $AC+AB=9cm$

$AC=9-4=5cm$

So, $\cot C=\frac{BC}{AB}=\frac{3}{4},cosecC=\frac{AC}{AB}=-\frac{5}{4},\sec C=\frac{AC}{BC}=\frac{5}{3}$.

TRIGONOMETRY

Ex. 5 Given that $\cos (A-B)=\cos A\cos B+\sin B$, find the value of $\cos {15}^{\circ }$.

TRIGONOMETRY

Sol. Putting $A={45}^{\circ }$ and $B={30}^{\circ }$

We get $\cos ({45}^{\circ }-{30}^{\circ })=\cos {45}^{\circ }\cos {30}^{\circ }+\sin {45}^{\circ }\sin {30}^{\circ }$

$\Rightarrow \cos {15}^{0.}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$ $\Rightarrow \cos {15}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$

TRIGONOMETRY

Ex. 6 A Rhombus of side of $10cm$ has two angles of ${60}^{\circ }$ each. Find the length of diagonals and also find its area.

TRIGONOMETRY

Sol. Let $ABCD$ be a rhombus of side $10cm$ and $\mathrm{\angle }BAD=\mathrm{\angle }BCD={60}^{\circ }$. Diagonals of parallelogram bisect each other.

$S,AO=OC$ and $BO=OD$

In right triangle $AOB$

$\sin {30}^{\circ }=\frac{OB}{AB}\cos {30}^{\circ }=\frac{OA}{AB}$

$\Rightarrow \frac{1}{2}=\frac{OB}{10}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{OA}{10}$

$\Rightarrow OB=5cm\Rightarrow OA=5\sqrt{3}$

$\therefore BD=2(OB)\Rightarrow AC=2(OA)$

$\Rightarrow BD=2(5)\Rightarrow AC=2(5\sqrt{3})$

$\Rightarrow BD=10cm\Rightarrow AC=10\sqrt{3}cm$

TRIGONOMETRY

So, the length of diagonals $AC=10\sqrt{3}cm$&$BD=10cm$

Area of Rhombus $=\frac{1}{2}\times AC\times BD$

$=\frac{1}{2}\times 10\sqrt{3}\times 10=50\sqrt{3}c{m}^2.$

TRIGONOMETRY

Ex. 7 Evaluate : $\frac{{\sec }^2{54}^0-{\cot }^2{36}^{\circ }}{cose{c}^2{57}^0-{\tan }^2{33}^0}+2{\sin }^2{38}^{\circ }{\sec }^2{52}^{\circ }-{\sin }^2{45}^{\circ }+\frac{2}{\sqrt{3}}\tan {17}^0\tan {60}^{\circ }\tan {73}^{\circ }$

TRIGONOMETRY

Sol. $\frac{{\sec }^2{54}^0-{\cot }^2{36}^0}{cose{c}^2{57}^0-{\tan }^2{33}^0}+2{\sin }^2{38}^0{\sec }^2{52}^{\circ }-{\sin }^2{45}^{\circ }+\frac{2}{\sqrt{3}}\tan {17}^0\tan {60}^0\tan {73}^0$

$=\frac{{\sec }^2({90}^0-{36}^{\circ })-{\cot }^2{36}^0}{cosec({90}^0-{33}^0)-{\tan }^2{33}^0}+2{\sin }^2{38}^0{\sec }^2({90}^{\circ }-{38}^0)-{\sin }^2{45}^{\circ }+\frac{2}{\sqrt{3}}\tan ({90}^{\circ }-{73}^0)\tan {73}^0\tan {60}^{\circ }$

$=\frac{cose{c}^2{36}^0-{\cot }^2{36}^0}{{\sec }^2{33}^0-{\tan }^2{33}^0}+2{\sin }^2{38}^{0}cose{c}^2{38}^0-(\frac{1}{\sqrt{2}}{)}^2+\frac{2}{\sqrt{3}}\cot {73}^0\tan {73}^0\times \sqrt{3}$

$=\frac{1}{1}+2{\sin }^2{38}^0\times \frac{1}{{\sin }^2{38}^0}-\frac{1}{2}+\frac{2}{\sqrt{3}}\times \frac{1}{\tan {73}^0}\times {73}^0\times \sqrt{3}[\because cose{c}^2\theta -{\cot }^2\theta =1,{\sec }^2\theta -{\tan }^2\theta =1]$

$=1+2-\frac{1}{2}+2=5-\frac{1}{2}=\frac{9}{2}$.

TRIGONOMETRY

Ex. 8 Prove that : $cosec({65}^{\circ }+\theta )-\sec ({25}^{\circ }-\theta )-\tan ({55}^{\circ }-\theta )+\cot ({35}^{\circ }+\theta )=0$

TRIGONOMETRY

Sol. $cosec({65}^{\circ }+\theta )=cosec{90}^0-({25}^0-\theta )=\sec ({25}^{\circ }-\theta )$ …(i)

$\cot ({35}^{\circ }+\theta )=\cot {90}^{\circ }-({55}^{\circ }-\theta )=\tan ({55}^{\circ }-\theta )$…(ii)

$\therefore$ L.H.S. $cosec({65}^{\circ }+\theta )-\sec ({25}^{\circ }-\theta )-\tan ({55}^{\circ }-\theta )+\cot ({35}^{\circ }+\theta )$

$=\sec ({25}^{\circ }-\theta )-\sec ({25}^{\circ }-\theta )-\tan ({55}^{\circ }-\theta )+\tan ({55}^{\circ }-\theta )$

$=0[using(i)$&$(ii)]$ R.H.S.

TRIGONOMETRY

Ex. 9 Prove that : $\cot \theta -\tan \theta =\frac{2{\cos }^2\theta -1}{\sin \theta \cos \theta }$

TRIGONOMETRY

Sol. L.H.S. $\cot \theta -\tan \theta$

$=\frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos \theta }$ $[\because \cot \theta =\frac{\cos \theta }{\sin \theta },\tan \theta =\frac{\sin \theta }{\cos \theta }]$

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

$=\frac{{\cos }^2\theta -1+{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{2{\cos }^2\theta -1}{\sin \theta \cos \theta }$ R.H.S. Hence Proved.

TRIGONOMETRY

Ex. 10 Prove that: $(cosesA-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1$.

TRIGONOMETRY

Sol. L.H.S. $(cosecA-\sin A)(\sec A-\cos A)(\tan A+\cot A)$

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A})$

$=(\frac{1-{\sin }^{2}A}{\sin A})(\frac{1-{\cos }^{2}A}{\cos A})(\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A})$

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

$=1$

R.H.S. Hence Proved

TRIGONOMETRY

Ex. 11 If $\sin \theta +\cos \theta =m$ and $\sec \theta +cosec\theta =n$, then prove that $n(m^2-1)=2m$.

TRIGONOMETRY

Sol. L.H.S. $n(m^2-1)$

$\begin{array}{c}=(\sec \theta +cosec\theta )[(\sin \theta +\cos \theta {)}^2-1]=(\frac{1}{\cos \theta }+\frac{1}{\sin \theta })({\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta -1)\\ =(\frac{\cos \theta +\sin \theta }{\sin \theta \cos \theta })(1+2\sin \theta \cos \theta -1)=\frac{(\cos \theta +\sin \theta )}{\sin \theta \cos \theta }(2\sin \theta \cos \theta )\end{array}$

$=2(\sin \theta +\cos \theta )=2m\text{R.H.S.}\text{ Hence Proved. }$

TRIGONOMETRY

Ex. 12 If $\sec \theta =x+\frac{1}{4x}$, then prove that $\sec \theta +\tan \theta =2x$ or $\frac{1}{2x}$.

TRIGONOMETRY

Sol. $\sec \theta =x+\frac{1}{4x}$ …(i)

$\begin{array}{rl}& \therefore 1+{\tan }^2\theta ={\sec }^2\theta \\ & \Rightarrow {\tan }^2\theta ={\sec }^2\theta -1\Rightarrow {\tan }^2\theta =(x+\frac{1}{4x}{)}^2-1\\ & \Rightarrow {\tan }^2\theta =x^2+\frac{1}{16x^2}+2\times x\times \frac{1}{4x}-1\\ & \Rightarrow {\tan }^2\theta =x^2+\frac{1}{16x^2}+\frac{1}{2}-1\Rightarrow \tan \theta =\pm (x-\frac{1}{4x})\\ & \Rightarrow {\tan }^2\theta =(x-\frac{1}{4x}{)}^2\Rightarrow \tan \theta =\pm (x-\frac{1}{4x})\\ & \text{ So, }\tan \theta =x-\frac{1}{4x}\dots (ii)\\ & \text{ or }\tan \theta =-(x-\frac{1}{4x})\dots (iii)\end{array}$

TRIGONOMETRY

Adding equation (i) and (ii)

$\sec \theta +\tan \theta =x+\frac{1}{4x}+x-\frac{1}{4x}$

$\sec \theta +\tan \theta =2x$

Adding equation (i) and (ii)

$\begin{array}{rl}& \sec \theta +\tan \theta =x+\frac{1}{4x}-x+\frac{1}{4x}\\ & =\frac{1}{2x}\text{ Hence, }\sec \theta +\tan \theta +2x\text{ or }\frac{1}{2x}.\end{array}$

Ex. 13 If $\theta$ is an acute angle and $\tan \theta +\cot \theta =2$ find the value of ${\tan }^9\theta +{\cot }^9\theta$

TRIGONOMETRY

Sol. We have, $\tan \theta +\cot \theta =2$

$\Rightarrow \tan \theta +\frac{1}{\tan \theta }=2$

$\Rightarrow {\tan }^2\theta +1=2\tan \theta$

$\Rightarrow (\tan \theta -1{)}^2=0$

$\Rightarrow \tan \theta =1$

$\therefore {\tan }^9\theta +{\cot }^9\theta$

$={\tan }^9{45}^0+{\cot }^9{45}^0$

$=(1{)}^9+(1{)}^9$

TRIGONOMETRY

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

$\Rightarrow {\tan }^2\theta -2\tan \theta +1=0$

$\Rightarrow \tan \theta -1=0$

$\Rightarrow \tan \theta =\tan {45}^0$ $\Rightarrow \theta ={45}^0$

$=(\tan 45{)}^9+(\cot 45{)}^0$

$=2\text{. }$

TRIGONOMETRY

DAILY PRACTIVE PROBLEMS 11

OBJECTIVE DPP - 11.1

1. If $\alpha +\beta =\frac{\pi }{2}$ and $\alpha =\frac{1}{3}$, then $\sin \beta$ is

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

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

(C) $\frac{2}{3}$ $$

(D) $\frac{3}{4}$

TRIGONOMETRY

TRIGONOMETRY

2. If $5\tan \theta =4$, then value of $\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }$ is

(A) $\frac{1}{3}$ $$

(B) $\frac{1}{6}$ $$

(C) $\frac{4}{5}$ $$

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

TRIGONOMETRY

TRIGONOMETRY

3. If $7\sin \alpha =24\cos \alpha ;0<\alpha <\frac{\pi }{2}$, then value of $14\tan \alpha -75\cos \alpha -7\sec \alpha$ is equal to

(A) 1 $$

(B) 2 $$

(C) 3 $$

(D) 4

TRIGONOMETRY

TRIGONOMETRY

4. Given $3\beta +5\cos \alpha ;\beta =5$, then the value of $(3\cos \beta -5\sin \beta {)}^2$ is equal to

(A) 9 $$

(B) $\frac{9}{5}$ $$

(C) $\frac{1}{3}$ $$

(D) $\frac{1}{9}$

TRIGONOMETRY

TRIGONOMETRY

5. If $\tan \theta =4$, then $(\frac{\tan \theta }{\frac{{\sin }^3\theta }{\cos \theta }+\sin \theta \cos \theta })$ is equal to

(A) 0 $$

(B) $2\sqrt{2}$ $$

(C) $\sqrt{2}$ $$

(D) 1

TRIGONOMETRY

TRIGONOMETRY

6. The value of $\tan 5^{\circ }\tan {10}^{\circ }\tan {15}^{\circ }{20}^{\circ }$ $\tan {85}^{\circ }$, is

(A) 1 $$

(B) 2 $$

(C) 3 $$

(D) None of these

TRIGONOMETRY

TRIGONOMETRY

7. As $x$ increases from 0 to $\frac{\pi }{2}$ the value of $\cos x$

(A) increases $$

(B) decreases $$

(C) remains constant $$

(D) increases, then decreases

TRIGONOMETRY

TRIGONOMETRY

8. Find the value of $x$ from the equation $x\sin \frac{\pi }{6}{\cos }^2\frac{\pi }{4}=\frac{{\cot }^2\frac{\pi }{6}\sec \frac{\pi }{3}\tan \frac{\pi }{4}}{cose{c}^2\frac{\pi }{4}cosec\frac{\pi }{6}}$

(A) 4 $$

(B) 6 $$

(C) - 2 $$

(D) 0

TRIGONOMETRY

TRIGONOMETRY

9. The area of a triangle is $12sq.cm$. Two sides are $6cm$ and $12cm$. The included angle is

(A) ${\cos }^{-1}(\frac{1}{3})$ $$

(B) ${\cos }^{-1}(\frac{1}{6})$ $$

(C) ${\sin }^{-1}(\frac{1}{6})$ $$

(D) ${\sin }^{-1}(\frac{1}{3})$

TRIGONOMETRY

TRIGONOMETRY

10. If $\alpha +\beta ={90}^{\circ }$ and $\alpha =2\beta$ then ${\cos }^2\alpha +{\sin }^2\beta$ equals to

(A) $\frac{1}{2}$ $$

(B) 0 $$

(C) 1 $$

(D) 2

TRIGONOMETRY

TRIGONOMETRY

SUBJECTIVE DPP - 11.2

1. Evaluate : (A) $\frac{\sin \theta \cos \theta \sin ({90}^{\circ }-\theta )}{\cos ({90}^{\circ }-\theta )}+\frac{\cos \theta \sin \theta \cos ({90}^{\circ }-\theta )}{\sin ({90}^{\circ }-\theta )}+\frac{{\sin }^2{27}^{\circ }+{\sin }^2{63}^{\circ }}{{\cos }^2{40}^{\circ }+{\cos }^2{50}^{\circ }}$

(B) $\cos {10}^{\circ }\cos 2^{\circ }\cos 3^{\circ }——–\cos {180}^{\circ }$

(C) $\sin ({50}^{\circ }+\theta )-\cos ({40}^{\circ }-\theta )+\tan 1^0\tan {20}^{\circ }\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }$

(D) $\frac{2}{3}({\cos }^4{30}^{\circ }-{\sin }^4{45}^{\circ })-3({\sin }^2{60}^{\circ }-{\sec }^2{45}^{\circ })+\frac{1}{4}{\cot }^2{30}^0$

(E) $\frac{{\cos }^2{20}^{\circ }+{\cos }^2{70}^{\circ }}{{\sec }^{2}50-{\cot }^2{40}^{\circ }}+2cosec{}^2{58}^{\circ }-2\cot {58}^{\circ }\tan {32}^{\circ }-4\tan {13}^0\tan {37}^{\circ }\tan {45}^{\circ }\tan {53}^{\circ }\tan {77}^{\circ }$

TRIGONOMETRY

TRIGONOMETRY

2. If $\cot \theta =\frac{3}{4}$, prove that $\sqrt{\frac{\sec \theta -cosec\theta }{\sec \theta +cosec\theta }}=\frac{1}{\sqrt{7}}$.

TRIGONOMETRY

3. If $A+B={90}^{\circ }$, prove that : $\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{{\sin }^{2}B}{{\cos }^{2}A}}=\tan A$

TRIGONOMETRY

4. If $A,B,C$ are the interior angles of a $\mathrm{△}ABC$, show that :

(i) $\sin \frac{B+C}{2}=\cos \frac{A}{2}$ $$ (ii) $\cos \frac{B+C}{2}=\sin \frac{A}{2}$

TRIGONOMETRY

Prove the following (Q, 5 to Q. 13)

5. ${\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta {\sin }^2\theta$

TRIGONOMETRY

6. $(\sin \theta +cosec\theta )+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$

[CBSE - 2008]

TRIGONOMETRY

7. $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=\sec \theta cosec\theta +1$

TRIGONOMETRY

8. $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sec \theta -\tan \theta$

TRIGONOMETRY

9. $\frac{\sin A+\cos A}{\sin A+\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{{\sin }^{2}A-{\cos }^{2}A}=\frac{2}{1-2{\cos }^{2}A}$

TRIGONOMETRY

10. $(\sin \theta +\sec \theta {)}^2+(\cos \theta +cosec\theta {)}^2=(1+\sec \theta$ $cosec\theta {)}^2$

TRIGONOMETRY

11. $(1+\cot \theta -cosec\theta )(1+\tan \theta +\sec \theta )=2$

TRIGONOMETRY

12. $({\sin }^8\theta -{\cos }^8\theta )=({\sin }^2\theta -{\cos }^2\theta )(1-2{\sin }^2\theta {\cos }^2\theta )$

TRIGONOMETRY

13. $\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{1+\sin \theta }{\cos \theta }$

TRIGONOMETRY

14. If $x=r\sin \theta \cos \varphi ,y=r\sin \theta \sin \varphi ,z=r\cos \theta$, then Prove that : $x^2+y^2+z^2=r^2$.

TRIGONOMETRY

15. If $\cot \theta +\tan \theta =x$ and $\sec \theta -\cos \theta =y$, then prove that $(x^{2}y{)}^{2/3}-(x{y}^2{)}^{2/3}=1$

TRIGONOMETRY

16. If $\sec \theta +\tan \theta =p$, then show that $\frac{p^2-1}{p^2+1}=\sin \theta$

[CBSE - 2004]

TRIGONOMETRY

17. Prove that : ${\tan }^{2}A-{\tan }^{2}B=\frac{{\cos }^{2}B-{\cos }^{2}A}{{\cos }^{2}B{\cos }^{2}A}=\frac{{\sin }^{2}A-{\sin }^{2}B}{{\sin }^{2}A{\sin }^{2}B}$

[CBSE - 2005]

TRIGONOMETRY

18. Prove that : $\frac{1}{\sec x-\tan x}-\frac{1}{\cos x}=\frac{1}{\cos x}=\frac{1}{\cos x}-\frac{1}{\sec x+\tan x}$

[CBSE - 2005]

TRIGONOMETRY

19. Prove : $(1+{\tan }^{2}A)+(1+\frac{1}{{\tan }^{2}A})=\frac{1}{{\sin }^{2}A-{\sin }^{4}A}$

[CBSE - 2006]

TRIGONOMETRY

20. Evaluate :

$\tan 7^{\circ }\tan {23}^{\circ }\tan {60}^{\circ }\tan {67}^{\circ }\tan {83}^{\circ }+\frac{\cot {54}^{\circ }}{\tan {36}^{\circ }}+\sin {20}^{\circ }\sec {70}^{\circ }-2$.

[CBSE - 2007]

TRIGONOMETRY

TRIGONOMETRY

21. Without using trigonometric tables, evaluate the following :

$({\sin }^2{65}^{\circ }+{\sin }^2{25}^{\circ })+\sqrt{3}(\tan 5^{\circ }\tan {15}^{\circ }\tan {30}^{\circ }\tan {75}^{\circ }\tan {85}^{\circ })$

[CBSE - 2008]

TRIGONOMETRY

TRIGONOMETRY

22. If $\sin 3\theta =\cos (\theta -{60}^{\circ })$ and $3\theta$ and $\theta -{60}^{\circ }$ are acute, find the value of $\theta$

[CBSE - 2008]

TRIGONOMETRY

TRIGONOMETRY

23. If $\sin \theta =\cos \theta$, find the value of $\theta$.

[CBSE - 2008]

TRIGONOMETRY

TRIGONOMETRY

24. If $7{\sin }^2\theta +3{\cos }^2\theta =4$, show that $\tan \theta =\frac{1}{\sqrt{3}}$

[CBSE - 2008]

TRIGONOMETRY

25. Prove : $\sin \theta (1+\tan \theta )+\cos \theta (1+\cot \theta )=\sec \theta +cosec\theta$.