Important Questions For Class 11 Maths- Chapter 3 - Trigonometric Functions

Trigonometric Functions is one of the most important topics of Maths. It is an important topic for joint entrance exams and also for class 12(CBSE). Apart from studying and practicing problems on trigonometric functions from NCERT, students shall also practice these important questions.

Solving these important questions of class 11 maths chapter 3 will help you prepare for joint entrance exams too.

Question 1: Prove the following identity : sec4 a − sec2 a = tan4 a + tan2 a

Solution:

LHS = sec4 a – sec2 a

= sec2 a (sec2 a-1)

= (tan2 a +1)(tan2 a) (Therefore, sec2 a – tan2 a = 1)

= tan4 a+tan2 a which is RHS

Hence proved.

Question 2: In a quadrilateral PQRS, prove that sin(P+Q) + sin(R + S) =0

Solution:

P + Q+ R+ S = 2 π

P + Q = 2π – (R+S)

sin (P+Q) = sin( 2π – (R+S))

sin (P+Q) = -sin(R+S)

sin (P+Q) – sin (R+S) =0

Question 3: In a quadrilateral PQRS, prove that cos(P + Q) = cos(R+S)

Soln: P + Q+ R+ S = 2 π

P + Q = 2π – (R+S)

cos (P+Q) = cos( 2π – (R+S))

cos (P+Q) = cos(R+S)

Check out more: Trigonometric Identities, Trigonometry Table & Sum and Difference of Angles – Trigonometry

Question 4: If P, Q, R, S are angles of a cyclic quadrilateral, prove that cos A + cos B + cos C + cos D = 0.

Solution:

= P + R = π and Q + S = π

So, P = π – R and Q = π – S

= Cos P = cos(π – R) = -Cos R and Cos Q = Cos(π – S) = – Cos S

= Cos P + Cos Q + Cos R + Cos S = -Cos R – Cos S + Cos R + Cos S = 0

Question 5: Prove the following identity: (1 + cot a – cosec a) ( 1 + tan a + sec a) = 2

Solution:

= (1 + cot a – cosec a) (1 + tan a + sec a)

= (1+ (cosa/sina) – (1/sina)) (1 +(sina/cosa) + (1/cosa))

= [(sin a + cos a – 1) (sina + cos a + 1)] / sin a cos a

= [(sin a + cos a )2 – 1] /sina cosa

= [Sin2a + cos2a + 2sinacosa -1] / sin a cos a

= 2sin a cos a/ sin a cos a = 2

Question 6: If cos a + sin a = √ 2 cos a, prove that cos a – sin b = √ 2 sin a

Solution:

= (cos a + sin a)2 + (cos a – sin a)2 = 2

= (√ 2 cos a)2 + (cos a – sin a)2 = 2

= (cos a – sin a)2 = 2 – (√ 2 cos a)2

= (cos a – sin a)2 = 2 sin2a = √ 2sin a

Question 7: Prove that sin 2 π/6 + cos2 π/3 – tan2 π/4 = -½

Solution:

sin 2 π/6 + cos 2 π/3 – tan 2 π/4

= (sin π/6)2 + (cosπ/3)2 – (tan π/4)2

= (½)2 + (½)2 – 1

= ¼ + ¼ -1 = -½

Question 8: Evaluate sin (7π/4)

Solution:

As we can see that sin (7π/4) is in the fourth quadrant, where sine function is negative. And 3(an odd number)

sin (7π/4) = sin(3 * π/2 + π/4) = -cos π/4 = -1/√ 2

Question 9: Prove the identity, (sina +cosec a)2 + (cos a + sec a)2 = tan 2 a + cot 2 a + 7

Solution:

= (sina +cosec a)2 + (cos a + sec a)2

= Sin2a + cosec2 a + 2sina cosec a +cos2a + sec2a + 2 cosa seca

= (Sin2a +cos2a) + (cosec2 a + sec2a) + 2 + 2

=1+ (1 + cot2a) + (1 + tan 2a) +4 = tan2 a + cot 2a + 7 = RHS

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