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

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Experts suggest students practice the solutions many number of times to yield 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.

## Download the PDF of 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.

### Access other exercises of RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions

Exercise 4.1 Solutions

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Exercise 4.4 Solutions

Exercise 4.5 Solutions

Exercise 4.6 Solutions

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Exercise 4.9 Solutions

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