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

 

  • 2tan245+co230sin260

 

  • cos45(sec30+cosec30)

 

  • (sin30+tan45cosec60)(sec30+cos60+cot45)

 

  • (5cos260+4sec230tan245)(sin230+cos230)

 

Ans.- (i) sin60cos30+sin30cos60

= (32×32)+(12×12)=34+14=44=1

 

(ii) 2tan245+co230sin260

=2×(1)2+(32)2(32)2=2

 

(iii) cos45(sec30+cosec30)

= 1223+2=12(2+23)3

= 32×(2+23)=322+26

 

= 3(2622)(26+22)(2622)

 

= 23(62)(262 (22)2)

 

23(62)248=23(62)16

 

3(62)8=(186)8=(326)8

 

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

= (12+12323+12+1)

= (322332+23)

= (334)2(33)242

= (27+16243)(2716)

= (43243)11

 

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

= 5(12)2+4(23)212(12)2+(32)2

= (54+1631)(14+34)

= (15+6412)1244

=6712

 

2) Find the correct answer and explain your choice:

 (i)  2tan301+tan230 =

          (A) sin 60 (B) cos 60 (C) tan 60 (D)        sin 30

 

 (ii) 1tan2451+tan230 =

tan 90 (B) 1  (C) sin 45  (D) 0

 

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

0 (B)  30    (C)  45   (D)  60

 

(iv)    2tan301tan230 =

cos 60 (B)  sin 60   (C)  tan 60     (D)  sin 30   

 

Ans.-

(i)  (A) IS correct.

2tan301+tan230 = 2(1)31+(13)2

(23)1+13=(23)43 =643=32=sin60

 

(ii)(D) is correct

1tan2451+tan230

= (112)(1+12)=02=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)

 

(23)113=2323=3=tan60

 

3) If tan (P + Q) = 3 and tan ( P – Q) = 13;00<P+Q<=90;P>Q
, calculate P and Q

                Ans:-     tan (P + Q) = 3

=>tan (P + Q) = tan 60°

=> (P + Q) =  60°     … (i)

=>tan (P – Q) = 13

=>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

 

(ii) True

Sin 0° = 0

Sin 30° = 12

Sin 45° = 12

Sin 60° = 32

Sin 90° = 1

Thus, the value of sinθ increases as θ increases

 

(iii) False

Cos 0° = 1

Cos 30° = 32

Cos 45° = 12

Cos 60° = 12

Cos 90° = 0

Thus, the value of Cosθ decreases as θ increases.

(iv) True

cotP=cosPSinP cot0=cos0Sin0=10=notdefined

 

Related Links
Ncert Books For Class 11 Chemistry Ncert Maths Book Class 11 Solutions Pdf
Ncert Exemplar Class 12 Chemistry Pdf Download Ncert Solutions Class 7 Maths
Ncert Solutions Class 8 Maths Ncert Solutions Class 9 Maths
Ncert 12 Biology Book Ncert Solutions Class 12 Maths
Ncert Books Class 6 Maths Ncert Grade 9 Science Solutions
Ncert Books For Class 9 Ncert Books For Class 6