Question

If 3 cos θ = 2 then (2 sec²θ + 2 tan²θ – 7) = ?

पर्याय

• 0

• 1

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Answer 1

0

Explanation:

`cos θ = 2/3`

⇒ `(AB)/(AC) = 2/3`

Let AB = 2k and AC = 3k. Then,

BC² = (AC² – AB²) 

= (9k² – 4k²) 

= 5k²

∴ `BC = sqrt(5)k`

`2(sec^2θ + tan^2θ) - 7 = 2[((AC)/(AB))^2 + ((BC)/(AB))^2] - 7`

= `2[((3k)/(2k))^2 + ((sqrt(5)k)/(2k))^2] - 7`

= `{2(3/2)^2 + 2((sqrt(5))/2)^2 - 7}`

= `(2 xx 9/4 + 2 xx 5/4 - 7)`

= 0