# Ncert Solutions For Class 10 Maths Ex 8.2

## Ncert Solutions For Class 10 Maths Chapter 8 Ex 8.2

1) Calculate the following:

• sin60cos30+sin30cos60$sin60^{\circ}cos30^{\circ}+sin30^{\circ}cos60^{\circ}$

• 2tan245+co230sin260$2tan^{2}45^{\circ}+co^{2}30^{\circ}-sin^{2}60^{\circ}$

• cos45(sec30+cosec30)$\frac{cos45^{\circ}}{\left ( sec30^{\circ}+cosec30^{\circ} \right )}$

• (sin30+tan45cosec60)(sec30+cos60+cot45)$\frac{\left ( sin30^{\circ}+tan45^{\circ}-cosec60^{\circ} \right )}{\left ( sec30^{\circ}+cos60^{\circ}+cot45^{\circ} \right )}$

• (5cos260+4sec230tan245)(sin230+cos230)$\frac{\left ( 5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2} 45^{\circ}\right )}{\left ( sin^{2} 30^{\circ}+cos^{2}30^{\circ}\right )}$

Ans.- (i) sin60cos30+sin30cos60$sin60^{\circ}cos30^{\circ}+sin30^{\circ}cos60^{\circ}$

= (32×32)+(12×12)=34+14=44=1$\left ( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right )+\left ( \frac{1}{2}\times \frac{1}{2} \right )=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$

(ii) 2tan245+co230sin260$2tan^{2}45^{\circ}+co^{2}30^{\circ}-sin^{2}60^{\circ}$

=2×(1)2+(32)2(32)2=2$2\times \left ( 1 \right )^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}-\left( \frac{\sqrt{3}}{2} \right )^{2}=2$

(iii) cos45(sec30+cosec30)$\frac{cos45^{\circ}}{\left ( sec30^{\circ}+cosec30^{\circ} \right )}$

= 1223+2=12(2+23)3$\frac{1}{\frac{\sqrt{2}}{2\sqrt{3}+2}}=\frac{1}{\frac{\sqrt{2}}{\frac{\left ( 2+2\sqrt{3} \right )}{\sqrt{3}}}}$

= 32×(2+23)=322+26$\frac{\sqrt{3}}{\sqrt{2}\times \left ( 2+2\sqrt{3} \right )}=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}$

= 3(2622)(26+22)(2622)$\frac{\sqrt{} 3\left ( 2\sqrt{6} -2\sqrt{2}\right )}{\left ( 2\sqrt{6}+2\sqrt{2} \right )\left ( 2\sqrt{6} -2\sqrt{2}\right )}$

= 23(62)(262 (22)2)$\frac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{\left ( 2\sqrt{6}^{2}\ -\left ( 2\sqrt{2} \right )^{2}\right )}$

23(62)248=23(62)16$\frac{2\sqrt{3}\left ( \sqrt{6} -\sqrt{2}\right )}{24-8}=\frac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{16}$

3(62)8=(186)8=(326)8$\frac{\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{8}=\frac{\left ( \sqrt{18}-\sqrt{6} \right )}{8}=\frac{\left ( 3\sqrt{2}-\sqrt{6} \right )}{8}$

(iv)  (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)

= (12+12323+12+1)$\left (\frac{ \frac{1}{2} + 1 – \frac{2}{\sqrt{3}}}{2\sqrt{3} + \frac{1}{2} + 1} \right )$

= (322332+23)$\left ( \frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{3}{2} + \frac{2}{\sqrt{3}}} \right )$

= (334)2(33)242$\frac{\left ( 3\sqrt{3} – 4 \right )^{2}}{\left ( 3\sqrt{3} \right )^{2} – 4^{2}}$

= (27+16243)(2716)$\frac{\left ( 27 + 16 – 24\sqrt{3} \right )}{\left ( 27 – 16 \right )}$

= (43243)11$\frac{\left ( 43 – 24\sqrt{3} \right )}{11}$

(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)

= 5(12)2+4(23)212(12)2+(32)2$5\left ( \frac{1}{2} \right )^{2} + 4 \left ( \frac{2}{\sqrt{3}} \right )^{2} – \frac{1^{2}}{\left ( \frac{1}{2} \right )^{2} + \left ( \frac{\sqrt{3}}{2} \right )^{2} }$

= (54+1631)(14+34)$\frac{\left ( \frac{5}{4} + \frac{16}{3 – 1} \right )}{\left ( \frac{1}{4} + \frac{3}{4} \right )}$

= (15+6412)1244$\frac{\frac{\left ( 15 + 64 – 12 \right )}{12}}{\frac{4}{4}}$

=6712$\frac{67}{12}$

2) Find the correct answer and explain your choice:

(i)  2tan301+tan230$\frac{2tan30^{\circ}}{1+tan^{2}30^{\circ}}$ =

(A) sin 60$60^{\circ}$ (B) cos 60$60^{\circ}$ (C) tan 60$60^{\circ}$ (D)        sin 30$30^{\circ}$

(ii) 1tan2451+tan230$\frac{1-tan^{2}45^{\circ}}{1+tan^{2}30^{\circ}}$ =

tan 90$90^{\circ}$ (B) 1  (C) sin 45$45^{\circ}$  (D) 0

(iii) sin 2P = 2 sin P is true when P =

0$0^{\circ}$ (B)  30$30^{\circ}$    (C)  45$45^{\circ}$   (D)  60$60^{\circ}$

(iv)    2tan301tan230$\frac{2tan30^{\circ}}{1-tan^{2}30^{\circ}}$ =

cos 60$60^{\circ}$ (B)  sin 60$60^{\circ}$   (C)  tan 60$60^{\circ}$     (D)  sin 30$30^{\circ}$

Ans.-

(i)  (A) IS correct.

2tan301+tan230$\frac{2tan30^{\circ}}{1+tan^{2}30^{\circ}}$ = 2(1)31+(13)2$2\frac{\left ( 1 \right )}{\frac{\sqrt{3}}{1}}+\left ( \frac{1}{\sqrt{3}} \right )^{2}$

(23)1+13=(23)43$\frac{\left ( 2\sqrt{3} \right )}{1+\frac{1}{3}} = \frac{\left (\frac{2}{\sqrt{3}}\right )}{\frac{4}{3}}$ =643=32=sin60$= \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2} = sin 60^{\circ}$

(ii)(D) is correct

1tan2451+tan230$\frac{1-tan^{2}45^{\circ}}{1+tan^{2}30^{\circ}}$

= (112)(1+12)=02=0$\frac{\left ( 1 – 1^{2} \right )}{\left ( 1 + 1^{2} \right )} = \frac{0}{2} = 0$

(iii) (A) is correct

sin 2P = 2 sin P is true when

P = sin 2P = sin 0° = 0
2 sin P = 2sin 0° = 2×0 = 0

or,

sin 2P = 2sin PcosP

=>2sin PcosP = 2 sin P

=>2cos P = 2 =>cosP = 1

=>P = 0°

(iv) (C) is correct

2tan301tan230=2(131(13)2)$\frac{2tan30^{\circ}}{ 1 – tan^{2}30^{\circ}} = 2\left ( \frac{\frac{1}{\sqrt{3}}}{1 – \left ( \frac{1}{\sqrt{3}} \right )^{2}} \right )$

(23)113=2323=3=tan60$\frac{\left ( 2\sqrt{3} \right )}{ 1 – \frac{1}{3}} = \frac{2\sqrt{3}}{\frac{2}{3}} = \sqrt{3} = tan 60^{\circ}$

3) If tan (P + Q) = 3$\sqrt{3}$ and tan ( P – Q) = 13$\frac{1}{\sqrt{3} }$;00<P+Q<=90;P>Q$0^{0} < P + Q <= 90^{\circ}; P>Q$
, calculate P and Q

Ans:-     tan (P + Q) = 3$\sqrt{3}$

=>tan (P + Q) = tan 60°

=> (P + Q) =  60°     … (i)

=>tan (P – Q) = 13$\frac{1}{\sqrt{3}}$

=>tan (P – Q) = 30°

=> (P – Q) = 30°     … (ii)

Adding (i) and (ii), we get

P + Q + P – Q = 60° + 30°

2P = 90°

=> P = 45°

Putting the value of P in equation (i)

45° + Q = 60°

=> Q = 60° – 45° = 15°

Hence, P = 45° and Q = 15°

4) Check whether the given statements are true or false, also give a reason for your answer:

(i) sin (P + Q) = sin P + sin Q.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cotP is not defined for P = 0°.

Ans:-

(i) False

Let P = 30° and Q = 60°, then
sin (P + Q) = sin (30° + 60°) = sin 90° = 1 and,
sin P + sin Q = sin 30° + sin 60°

= 12+32=1+32$\frac{1}{2} + \frac{\sqrt{3}}{2} = 1 + \frac{\sqrt{3}}{2}$

(ii) True

Sin 0° = 0

Sin 30° = 12$\frac{1}{2}$

Sin 45° = 12$\frac{1}{\sqrt{2}}$

Sin 60° = 32$\frac{\sqrt{3}}{2}$

Sin 90° = 1

Thus, the value of sinθ$sin \theta$ increases as θ$\theta$ increases

(iii) False

Cos 0° = 1

Cos 30° = 32$\frac{\sqrt{3}}{2}$

Cos 45° = 12$\frac{1}{\sqrt{2}}$

Cos 60° = 12$\frac{1}{2}$

Cos 90° = 0

Thus, the value of Cosθ$Cos \theta$ decreases as θ$\theta$ increases.

(iv) True

cotP=cosPSinP$cot P = \frac{cos P}{ Sin P}$ cot0=cos0Sin0=10=notdefined$cot 0^{\circ} = \frac{cos 0^{\circ}}{ Sin 0^{\circ}} = \frac{1}{0} = not \: defined$