If tan θ = p q , show that p sin θ - q cos θ p sin θ + q cos θ = p 2 - q 2 p 2 + q 2 .

We have tan θ = p q Let a right △ A B C , in which ∠ A = θ ° and ∠ B = 90 ° tan θ = Perpendicular Base = B C A B = p q Let A B = q k , then B C = p k , where k is a positive number. According to Pythagoras theorem, A C 2 = A B 2 + B C 2 A C = A B 2 + B C 2 A C = q k 2 + p k 2 A C = q 2 k 2 + p 2 k 2 = k p 2 + q 2 Now, sin θ = Perpendicular Hypotenuse = B C A C = p k k p 2 + q 2 = p p 2 + q 2 . . . i and cos θ = Base Hypotenuse = A B A C = q k k p 2 + q 2 = q p 2 + q 2 . . . i i ∴ p sin θ - q cos θ = p × p p 2 + q 2 - q × q p 2 + q 2 = p 2 p 2 + q 2 - q 2 p 2 + q 2 = p 2 - q 2 p 2 + q 2 . . . i i i and p sin θ + q cos θ = p × p p 2 + q 2 + q × q p 2 + q 2 = p 2 p 2 + q 2 + q 2 p 2 + q 2 = p 2 + q 2 p 2 + q 2 . . . i v So, p sin θ - q cos θ p sin θ + q cos θ = p 2 - q 2 p 2 + q 2 p 2 + q 2 p 2 + q 2 [from equation i i i and i v ] = p 2 - q 2 p 2 + q 2 × p 2 + q 2 p 2 + q 2 = p 2 - q 2 p 2 + q 2 Thus, p sin θ - q cos θ p sin θ + q cos θ = p 2 - q 2 p 2 + q 2 .

If tan θ = m n , show that m sin θ - n cos θ m sin θ + n cos θ = m 2 - n 2 m 2 + n 2 .

We have tan θ = m n Let a right △ A B C , in which ∠ A = θ ° and ∠ B = 90 ° tan θ = Perpendicular Base = B C A B = m n Let A B = n k , then B C = m k , where k is a positive number. Using Pythagoras theorem, A C 2 = A B 2 + B C 2 A C = A B 2 + B C 2 A C = n k 2 + m k 2 A C = n 2 k 2 + m 2 k 2 = k m 2 + n 2 Now, sin θ = Perpendicular Hypotenuse = B C A C = m k k m 2 + n 2 = m m 2 + n 2 . . . i and cos θ = Base Hypotenuse = A B A C = n k k m 2 + n 2 = n m 2 + n 2 . . . i i ∴ m sin θ - n cos θ = m × m m 2 + n 2 - n × n m 2 + n 2 = m 2 m 2 + n 2 - n 2 m 2 + n 2 = m 2 - n 2 m 2 + n 2 . . . i i i and m sin θ + n cos θ = m × m m 2 + n 2 + n × n m 2 + n 2 = m 2 m 2 + n 2 + n 2 m 2 + n 2 = m 2 + n 2 m 2 + n 2 . . . i v So, m sin θ - n cos θ m sin θ + n cos θ = m 2 - n 2 m 2 + n 2 m 2 + n 2 m 2 + n 2 [from equation i i i and i v ] = m 2 - n 2 m 2 + n 2 × m 2 + n 2 m 2 + n 2 = m 2 - n 2 m 2 + n 2 Thus, m sin θ - n cos θ m sin θ + n cos θ = m 2 - n 2 m 2 + n 2 .

E M B I B E

New Mathematics for Class X>Chapter 8 - Introduction To Trigonometry>EXERCISE

H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Introduction To Trigonometry, Exercise 1: EXERCISE

Author:H K Dass, Rama Verma & Bhagwat Swarup Sharma

H K Dass Mathematics Solutions for Exercise - H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Introduction To Trigonometry, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 8: Introduction To Trigonometry, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. New Mathematics for Class X solutions are prepared by Experienced Embibe Experts.