Trigonometrical Functions Class 11 OP Malhotra Exe-4A ISC Maths Solutions Ch-4 Solutions. In this article you would learn about Definition and Formula Expressing Relations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Trigonometrical Functions Class 11 OP Malhotra Exe-4A ISC Maths Solutions Ch-4

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-4	Trigonometrical Functions
Writer	OP Malhotra
Exe-4(A)	Definition and Formula Expressing Relations.

Exercise- 4A Definition and Formula Expressing Relations

Trigonometrical Functions Class 11 OP Malhotra Exe-4A Solution.

Prove that :

Que-1: 1 – cos² θ = sin² θ.

Sol: L.H.S. = 1 – cos² θ = 1 – (x²/r²)
= (r²−x²)/r² = (x²+y²−x²)/r²
= y²/r² = sin² θ = R.H.S

Que-2: √(1-sin²θ) = cos θ

Sol: L.H.S = √(1sin²θ)
= √{1-(y²/r²)} = √{(r²-y²)/r²} = √(x²/r²)
= ± x/r = ± cos θ [∵ x = r cos θ]

Que-3: sec a√(1-sin²a) = 1

Sol: L.H.S = sec a√(1-sin²a)
= sec a√(cos²a) = sec a | cos a |
= sec a x cos a = 1 if a lies in first and IV th quadrant. [in this case cos a > θ]

Que-4: (sec²θ−1)/(tan²θ) = 1

Sol: LHS = (sec²θ−1)/(tan²θ) = 1
= {(r²/x²)-1}/{y²/x²}
= {r²-x}/y²
= y²/y² = 1 RHS.

Que-5: sec² θ – 1 – tan² θ = 0.

Sol: L.H.S = sec² θ – 1 – tan² θ
= (r²/x²) − 1 − (y²/x² = (r²−x²−y²)/x²
= (r²−r²)/x² = 0 = R.H.S.

Que-6: (sin²θ+cos²θ)/(sec²θ−tan²θ) = 1.

Sol: L.H.S = (sin²θ+cos²θ)/(sec²θ−tan²θ)
= {(y²/r²)+(x²/r²)}/{(r²/x²)−(y²/x²)}
= {(x²+y²)/r²}/{(r²−y²)/x²}
= {(r²/r²}/{(x²/x²)} = 1
= R.H.S [∵ r² = x² + y²]

Que-7: 1 – cos² θ – sin² θ = 0.

Sol: L.H.S = 1 – cos² θ – sin² θ
= 1 – (cos² θ + sin² θ) = 1 – 1 = 0
= R.H.S

Que-8: sin⁴ θ + sin² θ cos² θ = sin² 0.

Sol: L.H.S = sin⁴ θ + sin² θ cos² θ
= sin² θ (sin² θ + cos² θ)
= sin² θ = R.H.S.

Que-9: sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ = 1.

Sol: L.H.S = sin⁴ θ + 2 sin² θ cos² θ + cos² θ
= (sin² θ + cos² θ)² = 1² = 1
= R.H.S.

Que-10: cos² θ (cosec² θ – cot² θ) = cos² θ.

Sol: L.H.S = cos² θ (cosec² θ – cot² θ)
= cos² θ . 1    [∵ 1 + cot² θ = cosec² θ]
= cos² θ = R.H.S

Que-11: {sinθ cosecθ tanθ cotθ}/{sin²θ+cos²θ} = 1.

Sol: LHS = {sinθ cosecθ tanθ cotθ}/{sin²θ+cos²θ}
= {sinθ . (1/sinθ) . tanθ . (1/tanθ)}
= 1 = RHS.

Que-12: {sin²30°+cos²30°}/{sec²57°−tan²57°} = 1

Sol: L.H.S = {sin²30°+cos²30°}/{sec²57°−tan²57°}
= 1/1 = 1    [∵ sin² 9 + cos² 9 = 1 = sec² θ – tan² θ]
= R.H.S.

Que-13: cos A tan A = sin A.

Sol: L.H.S. = cos A tan A = (x/r) x (y/x)
= y/r = sin A
= R.H.S.

Que-14: sin⁴ A cosec² A + cos⁴ A sec² A = 1.

Sol: LHS = sin⁴ A cosec² A + cos⁴ A sec² A
= sin⁴ A . (1/sin² A) + cos⁴ A . (1/cos² A)
= sin² + cos² = 1
= RHS.

Que-15: sin² A cot² A + cos² A tan² A = 1.

Sol: L.H.S = sin² A cot² A + cos² A tan² A
= cos² A + sin² A = 1
= R.H.S

–: End Trigonometrical Functions Class 11 OP Malhotra Exe-4A ISC Math Ch-4 Solution :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions

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