SSC CPO SI and ASI Model Paper 3

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Question : 102 of 200

Marks: +1, -0

Find the value of

tan^10

θ –

cot^15

θ, if tanθ + cotθ = 2 and θ is an acute angle.

1
0
-1
$1/ 2$

Solution:

(2) tanθ + cotθ= 2

⇒ tanθ+ 1/{tan}= 2

⇒tan^2θ + 1 = 2tanθ

⇒(tanθ-1)^2 = 0

(∴(a - b)^2=a^2 +b^2 - 2ab)

Squaring both sides,
⇒ tanθ = 1
⇒θ= 45°

∴ tan^10θ– cot^15θ = {(tan45)°}^10– {(cot45)°}15

= 1 – 1 = 0

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