RD Sharma Solutions for Class 11 Maths Chapter 5 - Trigonometric Functions PDF

RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions EXERCISE 5.1 PAGE NO: 5.18 Prove the following identities: 1. sec4 x – sec2 x = tan4 x + tan2 x Solution: Let us consider LHS: sec4 x – s...

RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions EXERCISE 5.1 PAGE NO: 5.18 Prove the following identities: 1. sec4 x – sec2 x = tan4 x + tan2 x Solution: Let us consider LHS: sec4 x – sec2 x (sec2 x)2 – sec2 x By using the formula, sec2 θ = 1 + tan2 θ. (1 + tan2 x) 2 – (1 + tan2 x) 1 + 2tan2 x + tan4 x – 1 - tan2 x tan4 x + tan2 x = RHS ∴ LHS = RHS Hence proved. 2. sin6 x + cos6 x = 1 – 3 sin2 x cos2 x Solution: Let us consider LHS: sin6 x + cos6 x (sin2 x) 3 + (cos2 x) 3 By using the formula, a3 + b3 = (a + b) (a2 + b2 – ab) (sin2 x + cos2 x) [(sin2 x) 2 + (cos2 x) 2 – sin2 x cos2 x] By using the formula, sin2 x + cos2 x = 1 and a2 + b2 = (a + b) 2 – 2ab 1 × [(sin2 x + cos2 x) 2 – 2sin2 x cos2 x – sin2 x cos2 x 12 - 3sin2 x cos2 x 1 - 3sin2 x cos2 x = RHS ∴ LHS = RHS Hence proved. 3. (cosec x – sin x) (sec x – cos x) (tan x + cot x) = 1 Solution: Let us consider LHS: (cosec x – sin x) (sec x – cos x) (tan x + cot x) By using the formulas cosec θ = 1/sin θ; sec θ = 1/cos θ; tan θ = sin θ / cos θ; cot θ = cos θ / sin θ Now,  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions 1 = RHS ∴ LHS = RHS Hence proved. 4. cosec x (sec x – 1) – cot x (1 – cos x) = tan x – sin x Solution: Let us consider LHS: cosec x (sec x – 1) – cot x (1 – cos x) By using the formulas cosec θ = 1/sin θ; sec θ = 1/cos θ; tan θ = sin θ / cos θ; cot θ = cos θ / sin θ Now, By using the formula, 1 – cos2x = sin2x;  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions = RHS ∴ LHS = RHS Hence Proved. 5. Solution: Let us consider the LHS: By using the formula, cosec θ = 1/sin θ; sec θ = 1/cos θ; Now,  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions sin x = RHS ∴ LHS = RHS Hence Proved. 6. Solution: Let us consider the LHS: By using the formula, tan θ = sin θ / cos θ; cot θ = cos θ / sin θ Now,  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions By using the formula, a3 - b3 = (a - b) (a2 + b2 + ab) By using the formula, cosec θ = 1/sin θ, sec θ = 1/cos θ; cosec x × sec x + 1 sec x cosec x + 1 =RHS ∴ LHS = RHS Hence Proved. 7. Solution: Let us consider LHS: By using the formula a3 ± b3 = (a ± b) (a2 + b2∓ ab)  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions We know, sin2x + cos2x = 1. 1 - sinx cosx + 1 + sinx cosx 2 = RHS ∴ LHS = RHS Hence Proved. 8. (sec x sec y + tan x tan y)2 – (sec x tan y + tan x sec y)2 = 1 Solution: Let us consider LHS: (sec x sec y + tan x tan y)2 – (sec x tan y + tan x sec y)2 Expanding the above equation we get, [(sec x sec y)2 + (tan x tan y)2 + 2 (sec x sec y) (tan x tan y)] – [(sec x tan y)2 + (tan x sec y)2 + 2 (sec x tan y) (tan x sec y)] [sec2 x sec2 y + tan2 x tan2 y + 2 (sec x sec y) (tan x tan y)] – [sec2 x tan2 y + tan2 x sec2 y + 2 (sec2 x tan2 y) (tan x sec y)] sec2 x sec2 y - sec2 x tan2 y + tan2 x tan2 y - tan2 x sec2 y sec2 x (sec2 y - tan2 y) + tan2 x (tan2 y - sec2 y) sec2 x (sec2 y - tan2 y) - tan2 x (sec2 y - tan2 y) We know, sec2 x – tan2 x = 1. sec2 x × 1 – tan2 x × 1 sec2 x – tan2 x 1 = RHS ∴ LHS = RHS Hence proved. 9. Solution: Let us Consider RHS: RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions = LHS ∴ LHS = RHS Hence Proved. 10. Solution: Let us consider LHS: By using the formulas, 1 + tan2x = sec2x and 1 + cot2x = cosec2x  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions = RHS ∴ LHS = RHS Hence Proved. 11. Solution: Let us consider LHS: By using the formula, tan θ = sin θ / cos θ; cot θ = cos θ / sin θ Now,  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions By using the formula, a3 + b3 = (a + b) (a2 + b2- ab) We know, sin2 x + cos2 x = 1. 1 – 1 + sin x cos x Sin x cos x = RHS ∴ LHS = RHS Hence proved. 12. Solution: Let us consider LHS: By using the formula, cosec θ = 1/sin θ, sec θ = 1/cos θ;  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions = RHS ∴ LHS = RHS Hence proved. 13. (1 + tan α tan β) 2 + (tan α – tan β) 2 = sec2 α sec2 β Solution: Let us consider LHS: (1 + tan α tan β) 2 + (tan α – tan β) 2 1+ tan2 α tan2 β + 2 tan α tan β + tan2 α + tan2 β – 2 tan α tan β 1 + tan2 α tan2 β + tan2 α + tan2 β tan2 α (tan2 β + 1) + 1 (1 + tan2 β) (1 + tan2 β) (1 + tan2 α) We know, 1 + tan2 θ = sec2 θ So, sec2 α sec2 β = RHS ∴ LHS = RHS Hence proved.  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions EXERCISE 5.2 PAGE NO: 5.25 1. Find the values of the other five trigonometric functions in each of the following: (i) cot x = 12/5, x in quadrant III (ii) cos x = -1/2, x in quadrant II (iii) tan x = 3/4, x in quadrant III (iv) sin x = 3/5, x in quadrant I Solution: (i) cot x = 12/5, x in quadrant III In third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative. By using the formulas, tan x = 1/cot x = 1/(12/5) = 5/12 cosec x = -√(1 + cot2 x) = -√(1 + (12/5)2) = -√(25+144)/25 = -√(169/25) = -13/5 sin x = 1/cosec x = 1/(-13/5) = -5/13 cos x = - √(1 - sin2 x) = - √(1 – (-5/13)2) = - √(169-25)/169 = - √(144/169) = -12/13 sec x = 1/cos x = 1/(-12/13) = -13/12 ∴ sin x = -5/13, cos x = -12/13, tan x = 5/12, cosec x = -13/5, sec x = -13/12 (ii) cos x = -1/2, x in quadrant II In second quadrant, sin x and cosec x are positive. tan x, cot x, cos x, sec x are negative.  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions By using the formulas, sin x = √(1 – cos2 x) = √(1 – (-1/2)2) = √(4-1)/4 = √(3/4) = √3/2 tan x = sin x/cos x = (√3/2)/(-1/2) = -√3 cot x = 1/tan x = 1/-√3 = -1/√3 cosec x = 1/sin x = 1/(√3/2) = 2/√3 sec x = 1/cos x = 1/(-1/2) = -2 ∴ sin x = √3/2, tan x = -√3, cosec x = 2/√3, cot x = -1/√3 sec x = -2 (iii) tan x = 3/4, x in quadrant III In third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative. By using the formulas, sin x = √(1 – cos2 x) = - √(1-(-4/5)2) = - √(25-16)/25 = - √(9/25) = - 3/5 cos x = 1/sec x = 1/(-5/4) = -4/5 cot x = 1/tan x  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions = 1/(3/4) = 4/3 cosec x = 1/sin x = 1/(-3/5) = -5/3 sec x = -√(1 + tan2 x) = - √(1+(3/4)2) = - √(16+9)/16 = - √ (25/16) = -5/4 ∴ sin x = -3/5, cos x = -4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3 (iv) sin x = 3/5, x in quadrant I In first quadrant, all trigonometric ratios are positive. So, by using the formulas, tan x = sin x/cos x = (3/5)/(4/5) = 3/4 cosec x = 1/sin x = 1/(3/5) = 5/3 cos x = √(1-sin2 x) = √(1 – (-3/5)2) = √(25-9)/25 = √(16/25) = 4/5 sec x = 1/cos x = 1/(4/5) = 5/4 cot x = 1/tan x = 1/(3/4) = 4/3  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions ∴ cos x = 4/5, tan x = 3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3 2. If sin x = 12/13 and lies in the second quadrant, find the value of sec x + tan x. Solution: Given: Sin x = 12/13 and x lies in the second quadrant. We know, in second quadrant, sin x and cosec x are positive and all other ratios are negative. By using the formulas, Cos x = √(1-sin2 x) = - √(1-(12/13)2) = - √(1- (144/169)) = - √(169-144)/169 = -√(25/169) = - 5/13 We know, tan x = sin x/cos x sec x = 1/cos x Now, tan x = (12/13)/(-5/13) = -12/5 sec x = 1/(-5/13) = -13/5 Sec x + tan x = -13/5 + (-12/5) = (-13-12)/5 = -25/5 = -5 ∴ Sec x + tan x = -5 3. If sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 find the value of 8 tan x -√5 sec y. Solution: Given: sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 We know that, x is in second quadrant and y is in third quadrant. In second quadrant, cos x and tan x are negative.  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions In third quadrant, sec y is negative. By using the formula, cos x = - √(1-sin2 x) tan x = sin x/cos x Now, cos x = - √(1-sin2 x) = - √(1 – (3/5)2) = - √(1 – 9/25) = - √((25-9)/25) = - √(16/25) = - 4/5 tan x = sin x/cos x = (3/5)/(-4/5) = 3/5 × -5/4 = -3/4 We know that sec y = - √(1+tan2 y) = - √(1 + (1/2)2) = - √(1 + 1/4) = - √((4+1)/4) = - √(5/4) = - √5/2 Now, 8 tan x - √5 sec y = 8(-3/4) - √5(-√5/2) = -6 + 5/2 = (-12+5)/2 = -7/2 ∴ 8 tan x - √5 sec y = -7/2 4. If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x. Solution: Given: Sin x + cos x = 0 and x lies in fourth quadrant. Sin x = -cos x Sin x/cos x = -1 So, tan x = -1 (since, tan x = sin x/cos x) We know that, in fourth quadrant, cos x and sec x are positive and all other ratios are negative.  RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions By using the formulas, Sec x = √(1 + tan2 x) Cos x = 1/sec x Sin x = - √(1- cos2 x) Now, Sec x = √(1 + tan2 x) = √(1 + (-1)2) = √2 Cos x = 1/sec x = 1/√2 Sin x = - √(1 – cos2 x) = - √(1 – (1/√2)2) = - √(1 – (1/2)) = - √((2-1)/2) = - √(1/2) = -1/√2 ∴ sin x = -1/√2 and cos x = 1/√2 5. If cos x = -3/5 and π