Question

10(sinx)^2 +15(cos x)^2=6 then what is 27(cosec x)^2+8(sec x)^2=

Answer 1

10(sinx)² + 15(cosx)² = 6
⇒10sin²x + 15cos²x = 6
we know, sin²Φ + cos²Φ = 1 so, cos²Φ = 1 - sin²Φ

⇒ 10sin²x + 15(1 - sin²x) = 6
⇒ 10sin²x + 15 - 15sin²x = 6
⇒ -5sin²x = -9
⇒ sin²x = 9/5 ∴cosec²x = 1/sin²x = 5/9 -----(1)

Similarly , 10(1 - cos²x) + 15cos²x = 6 [ ∵ sin²Ф = 1 - cos²Ф by identity ]
⇒10 - 10cos²x + 15cos²x = 6
⇒ 5cos²x = -4
⇒cos²x = 4/5 ∴sec²x = 1/cos²x = 5/4 ----(2)

Now, 27(cosecx)² + 8(secx)² = 27cosec²x + 8sec²x
= 27 × 5/9 + 8 × 5/4 [ from equations (1) and (2)]
= 15 + 10 = 25

Hence, answer is 25