If A , B and C are interior angles of ∆ A B C , then show that tan 2 B + C 2 = cos e c 2 A 2 - 1 .

Given, A + B + C = 180 ο ⇒ B + C = 180 ο - A ⇒ B + C 2 = 90 ο - A 2 ⇒ tan 2 B + C 2 = tan 2 ( 90 ο - A 2 ) ⇒ tan 2 B + C 2 = c o t 2 A 2 ⇒ t an 2 B + C 2 = cos e c 2 A 2 - 1 [ cos e c 2 A - c o t 2 A = 1 ] Hence, LHS=RHS (Proved)

If sec θ = x + 1 4 x , then find the value of sec θ + tan θ

sec θ = x + 1 4 x . . . . . . . . . . . . . . . . . . . . ( i ) sec 2 θ = x + 1 4 x 2 = x 2 + 2 x × 1 4 x + 1 4 x 2 = x 2 + 1 2 + 1 16 x 2 + 1 2 - 1 2 sec 2 θ = x 2 - 1 2 + 1 16 x 2 + 1 sec 2 θ - 1 = x 2 - 1 2 + 1 16 x 2 sec 2 θ - tan 2 θ = 1 tan 2 θ = x - 1 4 x 2 tan θ = ± x - 1 4 x If , tan θ = + x - 1 4 x sec θ + tan θ = x + 1 4 x + x - 1 4 x = 2 x If , tan θ = - x - 1 4 x sec θ + tan θ = x + 1 4 x - x - 1 4 x = x - x + 1 4 x + 1 4 x = 2 4 x = 1 2 x Therefore, sec θ + tan θ = 2 x or 1 2 x

Prove that sec 2 θ + cosec 2 θ = tan θ + cot θ .

We have to prove that sec 2 θ + cosec 2 θ = tan θ + cot θ L . H . S = sec 2 θ + cosec 2 θ = 1 + tan 2 θ + 1 + cot 2 θ sec 2 θ - tan 2 θ = 1 and cosec 2 θ - 1 = cot 2 θ = tan 2 θ + cot 2 θ + 2 = tan 2 θ + cot 2 θ + 2 tan θ cot θ = tan θ + cot θ 2 = tanθ + cotθ = R . H . S Hence, we proved

If cos α cos β = m a n d cos α sin β = n , then show that m 2 + n 2 cos 2 α = m 2 n 2 .

Given, m = cos α cos β and n = cos α sin β ⇒ m 2 = cos 2 α cos 2 β and n 2 = cos 2 α sin 2 β ∴ m 2 + n 2 cos 2 α = cos 2 α cos 2 β + cos 2 α sin 2 β cos 2 α = cos 2 α 1 cos 2 β + 1 sin 2 β cos 2 α = cos 2 α cos 2 β + sin 2 β cos 2 β sin 2 β cos 2 α (Using the identity: cos 2 β + sin 2 β = 1 ) = cos 2 α 1 cos 2 β sin 2 β cos 2 α = cos 2 α cos 2 β cos 2 α sin 2 β = m 2 n 2 Therefore, m 2 + n 2 cos 2 α = m 2 n 2