Question

Question 8 Find the general solution of the equation (sec 2x)^2 = 1- tan 2x

Class X1 - Maths -Trigonometric Functions Page 78

Answer 1

sec²2x = 1 - tan2x
we know,
sec²A - tan²A = 1
sec²A = 1 + tan²A

(1 + tan²2x) = 1 - tan2x
1 + tan²2x = 1 - tan2x
tan²2x + tan2x = 0
tan2x( tan2x + 1) = 0
tan2x = 0 or -1

tan2x = 0
2x = nπ/2

again,
tan2x = -1
tan2x = -tan(π/4)
tan2x = tan(π - π/4)
tan2x = tan(3π/4)

we know,
tan∅ = tanA
∅ = nπ + A

tan2x = tan(3π/4)
2x = nπ + (3π/4)
x = nπ/2 + (3π/8)

Hence, the solutions are
x = nπ/2 or nπ/2 + (3π/8)