NCERT exemlar solution

Chapter 8 - Introduction to Trigonometry &
Its Equation NCERT Exemplar - Class 10




EXERCISE 8.1
Choose the correct answer from the given four options:
4
Q1. If cos A = , then the value of tan A is
5 4
3 3
(a) (b) (c) (d) 5
5 4 3 3
4 B 4x
Sol. (b): cos A =  
5 H 5x
P2 + B2 = H2 (By Pythagoras theorem)
 P2 + (4x)2 = (5x)2
 P2 = 25x2 – 16x2
 P2 = 9x2
 P = 3x
P 3x 3
 tan A =   , which verifies option (b).
B 4x 4
1
Q2. If sin A = , then the value of cot A is
2 1 3
(a) 3 (b) (c) (d) 1
3 2
1 P 1x
Sol. (a): sin A =  
2 H 2x
B2 + P2 = H2
 B + (1x)2 = (2x)2
2

 B2 = 4x2 – 1x2
 B2 = 3x2  B = 3x
B 3x
 cot A =  3
P 1x
Hence, right option is (a).
Q3. The value of the expression
cosec (75° + q) – sec (15° – q) – tan (55° + q) + cot (35° – q) is
3
(a) –1 (b) 0 (c) 1 (d)
2
Sol. (b): (75° + q) and (15° – q), are complements of each other. Similarly,
(55° + q) and (35° – q) are also complements.
So, cosec (75° + q) – sec (15° – q) – tan (55° + q) + cot (35° – q)
= cosec [90° – (15° – q)] – sec (15° – q) – tan (55° + q) + cot [90° – (55° + q)]




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, Chapter 8 - Introduction to Trigonometry &
Its Equation NCERT Exemplar - Class 10



= sec (15° – q) – sec (15° – q) – tan (55° + q) + tan (55° + q)
=0
Hence, right option is (b).
a
Q4. Given that sin q = , then cos q is equal to
b
b b b2  a2 a
(a) (b) (c) (d)
2
b a 2 a b b  a2
2

a P ax
Sol. (c): sin q =   [Given]
b H bx
By Pythagoras theorem,
B2 + P2 = H2
 B + (ax)­2­ = (bx)2
2

 B2 = b2x2 – a2x2
2 2 2
 B2 = x (b  a )
 B = x b2  a2
2 2
B x (b  a ) b2  a2
 
cos q =  ,
H bx b
which verifies the option (c).
Q5. If cos (a + b) = 0, then sin (a – b) can be reduced to
(a) cos b (b) cos 2 b (c) sin a (d) sin 2a
Sol. (b): cos (a + b) = 0 [Given]
 cos (a + b) = cos 90°
 a + b = 90°
 a = 90° – b
Now, sin (a – b) = sin (90° – b – b)
= sin (90° – 2b)
 sin (a – b) = cos 2b
Hence, verifies the option (b).
Q6. The value of (tan 1° tan 2° tan 3° … tan 89°) is
1
(a) 0 (b) 1 (c) 2 (d)
2
Sol. (b): (tan 1° tan 2° tan 3° ... tan 89°)
= (tan 1° tan 89°) (tan 2° tan 88°) (tan 3° tan 87°) … (tan 45° tan 45°)
= [tan 1° tan (90° – 1)] [tan 2° tan (90° – 2)] [tan 3° tan(90° – 3)]…
tan 45° tan (90° – 45°)
= tan 1° cot 1° tan 2° cot 2° tan 3° cot 3° … tan 45° cot 45°
1 1 1 tan 45°
= tan 1°  tan 2°. tan 3°  
tan 1° tan 2° tan 3° tan 45°
= 1  1  1  1. … 1  1
=1
Hence, verifies the option (b).




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, Chapter 8 - Introduction to Trigonometry &
Its Equation NCERT Exemplar - Class 10



Q7. If cos 9a = sin a and 9a < 90°, then the value of tan 5a is
1
(a) (b) 3 (c) 1 (d) 0
3
Sol. (c): cos 9a = sin a
 cos 9a = cos (90° – a)
 9a = 90° – a
 10a = 90°
 a = 9°
 tan 5a = tan 5  9° = tan 45° = 1
Hence, verifies the option (c).
Q8. If DABC is right angled at C, then the value of cos (A + B) is
1 3
(a) 0 (b) 1 (c) (d)
2 2
Sol. (a): A + B +C = 180°
(Angle sum property of a triangle)
 (A + B) + 90° = 180°
 A + B = 90°
 cos (A + B) = cos 90° = 0
Hence, verifies the option (a).
Q9. If sin A + sin2A = 1, then the value of the expression (cos2A + cos4 A) is
(a) 1 (b) 1 (c) 2 (d) 3
2
Sol. (a): sin A + sin2 A = 1 [Given]
 sin A = (1 – sin2 A)
 sin A = cos2 A
 sin2 A = cos4 A [Squaring both sides]
 1 – cos2 A = cos4 A [ sin2 A = 1 – cos2 A]
 1 = cos2 A + cos4 A
Hence, verifies the option (a).
1 1
Q10. Given that sin  a , cos
 b , then value of (a + b) is
2 2
(a) 0° (b) 30° (c) 60° (d) 90°
1
Sol. (d): sin a = [Given]
2
 sin a = sin 30°
 a = 30°
1
Also, cos b =
2
 cos b = cos 60°
 b = 60°
 a + b = 30° + 60° = 90°
Hence, verifies the option (d).




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