S.chand Class 8 Maths Solution Chapter 16 Area of a Trapezium and a Polygon Exercise 16 A

 Exercise 16 A

Question 1

Find the area of the following trapeziums :

	Height	Parallel sides		Height	Parallel sides
(a)	7 cm	8 cm and 10 cm	(b)	15 cm	10 cm and 12 cm
(b)	2 cm	2.2 cm and 3.5 cm	(d)	4 cm	15 cm and 7.8 cm

2. A garden is in the form of a trapezium whose parallel sides are $40 \mathrm{~m}$ and $22 \mathrm{~m}$ and the perpendicular distance between them is $12 \mathrm{~m}$. Find the area of the garden.

3. Two parallel sides of a trapezium are $85 \mathrm{~cm}$ and $63 \mathrm{~cm}$ and its area is $2664 \mathrm{~cm}^{2}$. Find its altitude.

4. Find the height of the trapezium, the sum of the lengths of whose bases is $50 \mathrm{~cm}$, and whose area is $500 \mathrm{~cm}^{2}$

5. Find the sum of the lengths of the bases of a trapezium whose altitude is $17 \mathrm{~cm}$ and whose area is $0.85 \mathrm{~m}^{2}$.

6. The area of a trapezium is $210 \mathrm{~cm}^{2}$ and its height is $14 \mathrm{~cm}$. If one of the parallel sides is double that of the other, find the two parallel sides.

7. The area of a trapezium is $300 \mathrm{~m}^{2}$. The perpendicular distance between the two parallel sides is $15 \mathrm{~m}$. If the difference of the parallel sides is $16 \mathrm{~m}$, find the length of the parallel sides.

8. The lengths of the parallel sides of a trapezium are in the ratio $3: 5$ and the distance between them is $10 \mathrm{~cm}$. If the area of trapezium is $120 \mathrm{~cm}^{2}$, find the lengths of its parallel sides.

9. Two parallel sides of an isosceles trapezium are $6 \mathrm{~cm}$ and $14 \mathrm{~cm}$ respectively. If the length of each non-parallel side is $5 \mathrm{~cm}$, find the area of the trapezium.

10. $A B C D$ is a trapezium of area $91 \mathrm{~cm}^{2} . C D$ is parallel to $A B$ and $C D$ is longer than $A B$ by $8 \mathrm{~cm}$. If the distance between $A B$ and $C D$ is $7 \mathrm{~cm}$, find $A B$ and $C D$.

11. Find the cost of watering a trapezoidal field whose parallel sides are $10 \mathrm{~m}$ and $25 \mathrm{~m}$ respectively, the perpendicular distance between them is $15 \mathrm{~m}$ and the rate of watering is $₹ 4$ per $\mathrm{m}^{2}$.

12. The parallel sides of a trapezium are $25 \mathrm{~cm}$ and $13 \mathrm{~cm}$, its non-parallel sides are equal, each being $10 \mathrm{~cm}$. Find the area of the trapezium.

13. In the figure, $A B$ and $D C$ are parallel sides of a trapezium $A B C D$ and $\angle A D C=90^{\circ}$. Given $A B=15 \mathrm{~cm}, C D=40 \mathrm{~cm}$ and diagonal $A C=41 \mathrm{~cm}$, calculate the area of trapezium $A B C D$.

[Hint. In $\triangle A D C, A D=\sqrt{A C^{2}-D C^{2}}=\sqrt{41^{2}-40^{2}}=9 \mathrm{~cm}$ Now, Area of trap, $\left.A B C D=\frac{1}{2} \times A D \times(A B+D C)\right]$

14. Find the area of the pentagonal field shown alongside. All dimensions are in metres.

S Chand Class 10 CHAPTER 16 TRIGONOMETRY Exercise 16 A

 Exercise 16 A

Question 1

Ans: $\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}=1$

$L \cdot H \cdot S=\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}$ $\left[\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right]$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta}=1=R \cdot \mathrm{H} \cdot \mathrm{S}$

Question 2

Ans: $\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta}$=1

L.H.S $=\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta} \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\frac{\cos ^{2} \theta}{\cos ^{2} \theta}=1=R \cdot H \cdot S$

Question 3

Ans: $\sin A \cdot \cot A=\cos A$

L.H.S=$\sin A \cdot \cot A$

$=\sin A \cdot \frac{\cos A}{\sin A}$  $\left[\cot A=\frac{\cos A}{\sin A}\right]$

$=\cos A=R . H .S$

Question 4

Ans: $\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta=1$

$L . H . S=\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta \quad\left[\frac{1}{\cos ^{2} \theta}=\sec ^{2} \theta\right]$

$=\sec ^{2} \theta-\tan ^{2} \theta$

$=1=R \cdot H \cdot S$

Question 5

Ans: $\tan ^{2} \cos ^{2} A=1-\cos ^{2} A$

$L \cdot H \cdot S=\tan ^{2} A \cos ^{2} A$ $\left[\tan ^{2} A=\frac{\sin ^{2} A}{\cos ^{2} A}\right]$

$=\frac{\sin ^{2} A}{\cos ^{2} A} \cos ^{2} A$

$=\sin ^{2} A=1-\cos ^{2} A=R \cdot H \cdot S$

Question 6

Ans: $\tan \theta=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$

 $\Rightarrow R . H . S=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$ $\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\frac{\sin \theta}{\sqrt{\cos ^{2} \theta}}$

$=\frac{\sin \theta}{\cos \theta}=\tan \theta=$ L.H.S

Question 7

Ans: $\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1}{1-\cos \theta}$

L.H.S $=\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1+\cos \theta}{1-\cos ^{2} \theta}\left[\sin ^{2} \theta=1-\cos ^{2} \theta\right]$

$=\frac{1+\cos \theta}{(1-\cos \theta)(1-\cos \theta)}\left[a^{2}-b^{2}=(a-b)\left(a+
 b\right)\right.$

$=\frac{1}{1-\cos \theta}=$ R.H.S

Question 8

Ans: $\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)=\cos ^{2} \theta \quad\left[\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

$L \cdot H \cdot S=\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)$ $\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right]$

$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \times \sin ^{2} \theta$

$=\cos ^{2} \theta$ = R.H.S

Question 9

Ans: $\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta$

L.H.S $=\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)$

$\begin{array}{ll}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \cos ^{2} \theta & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]}\end{array}$

$=\sin ^{2} \theta=R \cdot H \cdot S$

Question 10

Ans: $\left(1-\sin ^{2} \theta\right) \operatorname{sic}^{2} \theta=1$

L.H.S $=\left(1-\sin ^{2} \theta\right) \sec ^{2} \theta \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\cos ^{2} \theta \cdot \sec ^{2} \theta$

$=1=$ R.H.S

Question 11

Ans: 

$\begin{aligned}\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta &=1 \\ L \cdot H \cdot S=\left(1-\cos ^{2} \theta\right) & \operatorname{cosec}^{2} \theta \\ 1-\cos ^{2} \theta &=\sin ^{2} \theta \\ \operatorname{cosec} \theta &=\frac{1}{\sin \theta} \\ &=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta} \\ 1^{-} &=R \cdot H \cdot S \end{aligned}$

Question 12

Ans: $\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=1$ 

$L \cdot H \cdot S=\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}$

$=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta} \quad\left[1+\tan ^{2} \theta=\sec ^{2} \theta\right]$

$\doteq \sin ^{2} \theta+\cos ^{2} \theta$ $\left[\frac{1}{\sec ^{2} \theta}=\cos ^{2} \theta\right]$

$R \cdot H .S=1$

Question 13

Ans:  $\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}=1$

$L \cdot H S=\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}$

$\left[1+\cot ^{2} \theta=\cos \operatorname{sc}^{2} \theta\right]$

$=\cos ^{2} \theta+\frac{1}{\operatorname{cosec}^{2} \theta}$

$\left\{\frac{1}{\operatorname{cosec}}=\sin \theta\right.$

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

$R \cdot H \cdot S=1$

Question 14

Ans: $\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$

$L \cdot H \cdot S=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$

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

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

$R \cdot H \cdot S=1$

Question 15

Ans: $\left[\frac{\cos ^{2} A}{\sin ^{2} A}+1\right] \tan ^{2} A=\frac{1}{\cos ^{2} A}$

$L \cdot H \cdot S=\left[\frac{\cos ^{2} \cdot A}{\sin ^{2} A}+1\right] \tan ^{2} \theta$

$\frac{\cos A}{\sin A}=\cot A$

$=\left(\cot ^{2} A+1\right) \tan ^{2} \theta$

$\quad \cos ^{2} A+1=\operatorname{cosec}^{2} A$

$=\operatorname{cosec}^{2} A \times \tan ^{2} A$

$\quad \operatorname{cosec} A=\frac{1}{\sin A}$

$=\frac{1}{\sin ^{2} A} \times \frac{\sin ^{2} A}{\cos ^{2} A}$

$R \cdot H \cdot S=\frac{1}{\cos ^{2} A}$

Question 16

Ans: $\sin ^{2} \theta+\sin ^{2} \theta \cos ^{2} \theta=\sin ^{2} \theta .$

$\begin{aligned} L \cdot H \cdot S=& \sin ^{4} \theta+\sin ^{2} \theta \cos ^{2} \theta \\=& \sin ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right.\\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=& \sin ^{2} \theta \times 1 \\ R \cdot H \cdot S=& \sin ^{2} \theta \end{aligned}$

Question 17

Ans: $\sin ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta+\cos 4 \theta=1$

$\begin{aligned} \text { L.H.S }=& \sin ^{2} \theta+2 \sin ^{2} \theta(a)^{2} \theta+\cos ^{4} \theta \\ & a^{2}+2 a b+b^{2}=(a+b)^{2} \\=&\left(\sin ^{2} \theta\right)^{2}+2 \sin ^{2} \theta \cos ^{2} \theta+\left(\cos ^{2} \theta\right)^{2} \\=&\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2} \\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=&(1)^{2} \end{aligned}$

R.H.S= 1

Question 18

Ans: $\sin 4 A \operatorname{cosec}^{2} A+\cos 4 A \sec ^{2} A=1$

L.H.S $=\sin 4 A \operatorname{cosec}^{2} A+\cos ^{4} A \sec ^{2} A$

$=\sin 4 A \times \frac{1}{\sin ^{2} A}+\cos ^{4} A \times \frac{1}{\cos ^{2} A}$  $\left[\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$\therefore \sin ^{2} \theta+\cos ^{2} \theta=1$ $\left[\sec \theta=\frac{1}{\cos \theta}\right]$

R.H.S =1

Question 19

Ans: $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$

$\begin{aligned} L \cdot H \cdot S=& \sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A \\=& \sin ^{2} A \times \frac{\cos ^{2} A}{\sin ^{2} A}+\cos ^{2} A \times \frac{\sin ^{2} A}{\cos ^{2} A} \\ & \frac{\cos A}{\sin A}=\cot \theta, \frac{\sin \theta}{\cos \theta}=\tan \theta \\=& \cos ^{2} A+\sin ^{2} A \cdot=1 \\ \text { R.H.S }=& 1 \end{aligned}$

Question 20

Ans: $\tan \theta+\cot \theta=\sec \theta \cdot \operatorname{cosec} \theta \cdot .$

$L \cdot H \cdot S=\tan \theta+\cot \theta$

${\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right] }$

$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}$

$=\left[\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta}\right]$

$=\frac{1}{\cos \theta \sin \theta}$

${\left[\frac{1}{\cos \theta}=\sec \theta, \frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }$

$R \cdot H \cdot S=\operatorname{Sec} \theta \cdot \operatorname{cosec} \theta$

Question 21

Ans:  $(\tan A+\cot A) \sin A \cos A=1$

$L \cdot H \cdot S=(\tan A+\cot A)(\sin A \cos A)$

$\begin{aligned} & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta \doteq \frac{\cos \theta}{\sin \theta}\right] } \\=& {\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right] \sin A \cos A . } \\=& \frac{\sin ^{2} A+\cos ^{2} A}{\cos A \sin A} \times \sin A \cos A \\ & {\left[\sin ^{2} \theta+\operatorname{Cos}^{2} \theta=1\right] }\end{aligned}$

$=\frac{1}{\cos A \sin A} \times \sin A \cos A$

R.H.S = 1

Question 22

Ans:   $\begin{aligned} \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta & \\ L \cdot H \cdot S=& \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)} \\=& \frac{1-\sin ^{2} \theta+\cos \theta}{\sin \theta(1+\cos \theta)} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{\cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta} \end{aligned}$

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

$=\frac{\cos \theta}{\sin \theta}$

${\left[\frac{\cos \theta}{\sin \theta}=\cot \theta\right] }$

$R \cdot H \cdot S=\cot \theta$

Question 23

Ans: 

 $\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=2 \operatorname{cosec}^{2} \theta$

$\begin{aligned} \text { L.H.S } &=\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta} \\ &=(a+b)(a-b)=a^{2}-b^{2} \\ &=\frac{1+\cos \theta+1-\cos \theta}{(1-\cos \theta)(1+\cos \theta)}=\frac{2}{1-\cos ^{2} \theta} \end{aligned}$

$\begin{aligned} & {\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] } \\=& \frac{2}{\sin ^{2} \theta} . \\ & {\left[\frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }\end{aligned}$

$R \cdot H \cdot S=2 \operatorname{cosec}^{2} \theta$

Question 24

Ans:  $\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\tan ^{2} \theta .$

$L \cdot H \cdot S=\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\frac{1-\tan ^{2} \theta}{\frac{1}{\tan ^{2} \theta}-1}$ 

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

$=\tan ^{2}\theta=R.H.S$

Question 25

Ans:  $\frac{1}{\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$

$L \cdot H \cdot S=\frac{1}{\operatorname{Sec} \theta+\tan \theta}$

$=\frac{\sec \theta-\tan \theta}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$

$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta} \quad\left[\sec ^{2} \theta-\tan ^{2} \theta=1\right]$

$=\frac{\sec \theta-\tan \theta}{1}$

$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$

R.H.S=$ \frac{1-\sin \theta}{\cos \theta}$

Question 26

Ans: $(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)=1$

L.H.S $=(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)$

$\left[\operatorname{cosec} A=\frac{1}{\sin A}, \sec A=\frac{1}{\cos A}, \tan A=\frac{\sin A}{\cos A}\right.$

$\left.\cot A=\frac{\cos A}{\sin A}\right]$

$=\left[\frac{1}{\sin A}-\sin A\right]\left[\frac{1}{\cos A}-\cos A\right]\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right]$

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

$=\frac{\left[\sin ^{2} A+\cos ^{2} A=1\right]}{\sin A} \times \frac{\sin ^{2} A}{\cos A} \times \frac{1}{\sin A \cos A}$

$=\frac{\cos ^{2} A \sin ^{2} A}{\sin ^{2} A \cos ^{2} A}$

R.H.S $=1$

Question 27

Ans: $\frac{1+\sin \theta}{1-\sin \theta}=(\sec \theta+\tan \theta)^{2}$

$\begin{aligned} \text { L.H.S. } &=\frac{1+\sin \theta}{1-\sin \theta} \\ &=\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)} \\ &=\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta} \end{aligned}$

$\begin{aligned} & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta} \\=& {\left[\frac{1+\sin \theta}{\cos \theta}\right]^{2} } \\=& {\left[\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right]^{2} }\end{aligned}$

$\frac{1}{\cos \theta}=\sec \theta, \frac{\sin \theta}{\operatorname{con} \theta}=\tan \theta$

$R \cdot H \cdot S=(\sec \theta+\tan \theta)^{2}$

Question 28

Ans: $\frac{1+\cos \theta}{1-\cos \theta}=(\operatorname{cosec} \theta+\cot \theta)^{2}$

$\begin{aligned} L \cdot H \cdot S &=\frac{1+\cos \theta}{1-\cos \theta} \\ &=\frac{(1+\cos \theta)(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)} \cdot\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] \end{aligned}$

$=\frac{\left(1+\cos (\theta)^{2}\right.}{1-\cos ^{2} \theta}$

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

$=\left[\frac{1+\cos \theta}{\sin \theta}\right]^{2}$

$=\left[\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right]^{2}$

$\left(\frac{1}{\sin \theta}=\operatorname{cosec} \theta, \frac{\cos \theta}{\sin \theta}=\cot \theta\right)$

$R \cdot H \cdot S=(\cos \theta+\theta+\cot \theta)^{2}$

Question 29

Ans: $\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\operatorname{cosec} \theta+\cot \theta-\frac{1+\cos \theta}{\sin \theta}$

$L \cdot H \cdot S=\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{\cot A+\operatorname{cosec} \theta-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{\cot \theta+\operatorname{cosec} \theta+\left(\cot ^{2} \theta-\operatorname{cosec} 2 \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\cot \theta+\operatorname{cosec} \theta)+\cot \theta+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\cot \theta+\operatorname{cosec} \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\operatorname{cotc}+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta+1)}{(\cot \theta-\operatorname{cosec} \theta+1}$

$=\cot \theta+\operatorname{cosec} \theta$

$=\operatorname{cosec} \theta+\cot \theta$

$=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta} .$

$R \cdot H \cdot S=\frac{1+\cos \theta}{\sin \theta}$

Question 30

Ans: $\tan \theta+\cot \theta=2$

$(\tan \theta+\cot \theta)^{2}=4 \quad$ [Square both side]

$\tan ^{2} \theta+\cot ^{2} \theta+2 \tan \theta \cot \theta=4$

${[\tan \theta \cdot \cot \theta=1] }$

$\begin{aligned} \tan ^{2} \theta+\cot ^{2} \theta+2 \times 1 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta+2 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta &=4-2 \\ &=2 \\ \tan ^{2} \theta+\cot ^{2} \theta &=2 \end{aligned}$

Question 31

Ans:  $a^{2}+b^{2}=(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2}$

$=-\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \epsilon \cos \theta$

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

$=2\left(\sin ^{2} \theta+\left(\cos ^{2} \theta\right)\right.$

$=2 \times 1$

$=2$

S.chand books class 8 maths solution chapter 16 Linear Inequations exercise 16 A

EXERCISE 16 A

Match the inequalities with the graph

Question 1

$x \leq-2$

Sol :

Question 2

$-4 \leq x<0$

Sol :

Question 3

$-3<x \leq 4$

Sol :

Question 4

$0 \leq x \leq 5$

Sol :

Question 5

x>0

Sol :

Question 6

$x \leq 4$

Sol :

Question 7

x>-3

Sol :

Question 8

x<2 or x>2

Sol :