RD Sharma Solutions for Class 12 Maths Exercise 4.14 Chapter 4 Inverse Trigonometric Functions

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Experts suggest students practise the solutions to secure good results in their exams. In Exercise 4.14 of Chapter 4 Inverse Trigonometric Functions, we shall discuss the ASA congruence condition for two congruent triangles and their properties.

RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.14

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Access answers to Maths RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.14

Exercise 4.14 Page No: 4.115

1. Evaluate the following:

(i) tan {2 tan^-1 (1/5) – π/4}

(ii) Tan {1/2 sin^-1 (3/4)}

(iii) Sin {1/2 cos^-1 (4/5)}

(iv) Sin (2 tan ^-1 2/3) + cos (tan^-1 √3)

Solution:

(i) Given tan {2 tan^-1 (1/5) – π/4}

(ii) Given tan {1/2 sin^-1 (3/4)}

(iii) Given sin {1/2 cos^-1 (4/5)}

(iv) Given Sin (2 tan ^-1 2/3) + cos (tan^-1 √3)

2. Prove the following results:

(i) 2 sin^-1 (3/5) = tan^-1 (24/7)

(ii) tan^-1 ¼ + tan^-1 (2/9) = ½ cos^-1 (3/5) = ½ sin^-1 (4/5)

(iii) tan^-1 (2/3) = ½ tan^-1 (12/5)

(iv) tan^-1 (1/7) + 2 tan^-1 (1/3) = π/4

(v) sin^-1 (4/5) + 2 tan^-1 (1/3) = π/2

(vi) 2 sin^-1 (3/5) – tan^-1 (17/31) = π/4

(vii) 2 tan^-1 (1/5) + tan^-1 (1/8) = tan^-1 (4/7)

(viii) 2 tan^-1 (3/4) – tan^-1 (17/31) = π/4

(ix) 2 tan^-1 (1/2) + tan^-1 (1/7) = tan^-1 (31/17)

(x) 4 tan^-1(1/5) – tan^-1(1/239) = π/4

Solution:

(i) Given 2 sin^-1 (3/5) = tan^-1 (24/7)

Hence, proved.

(ii) Given tan^-1 ¼ + tan^-1 (2/9) = ½ cos^-1 (3/5) = ½ sin^-1 (4/5)

Hence, proved.

(iii) Given tan^-1 (2/3) = ½ tan^-1 (12/5)

Hence, proved.

(iv) Given tan^-1 (1/7) + 2 tan^-1 (1/3) = π/4

Hence, proved.

(v) Given sin^-1 (4/5) + 2 tan^-1 (1/3) = π/2

(vi) Given 2 sin^-1 (3/5) – tan^-1 (17/31) = π/4

(vii) Given 2 tan^-1 (1/5) + tan^-1 (1/8) = tan^-1 (4/7)

Hence, proved.

(viii) Given 2 tan^-1 (3/4) – tan^-1 (17/31) = π/4

Hence, proved.

(ix) Given 2 tan^-1 (1/2) + tan^-1 (1/7) = tan^-1 (31/17)

Hence, proved.

(x) Given 4 tan^-1(1/5) – tan^-1(1/239) = π/4

Hence, proved.

3. If sin^-1 (2a/1 + a²) – cos^-1(1 – b²/1 + b²) = tan^-1(2x/1 – x²), then prove that x = (a – b)/ (1 + a b)

Solution:

Given sin^-1 (2a/1 + a²) – cos^-1(1 – b²/1 + b²) = tan^-1(2x/1 – x²)

Hence, proved.

4. Prove that:

(i) tan^-1{(1 – x²)/ 2x)} + cot^-1{(1 – x²)/ 2x)} = π/2

(ii) sin {tan^-1 (1 – x²)/ 2x) + cos^-1 (1 – x²)/ (1 + x²)} = 1

Solution:

(i) Given tan^-1{(1 – x²)/ 2x)} + cot^-1{(1 – x²)/ 2x)} = π/2

Hence, proved.

(ii) Given sin {tan^-1 (1 – x²)/ 2x) + cos^-1 (1 – x²)/ (1 + x²)}

Hence, proved.

5. If sin^-1 (2a/ 1+ a²) + sin^-1 (2b/ 1+ b²) = 2 tan^-1 x, prove that x = (a + b/ 1 – a b)

Solution:

Given sin^-1 (2a/ 1+ a²) + sin^-1 (2b/ 1+ b²) = 2 tan^-1 x

Hence, proved.

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