If tan θ – cot θ = 0 and θ is positive acute angle, then the value of \frac{\tan (\\theta+15^\circ)}{\tan (\\theta-15^\circ)} is

SSC CGL Tier II 25 Oct 2015 Paper 1

Directions (5–9) : The following graph shows production (in thousands) of two types (P and Q) of vehicles by a factory over the years 2009 to 2014. Study the graph and answer the given questions.

Question : 13 of 100

Marks: +1, -0

If tan θ – cot θ = 0 and θ is positive acute angle, then the value of

{\tan (\θ+15°)}/{\tan (\θ-15°)}

3
$1/√3$
$1/3$
$√3$

Validate

Solution:

(1) tanθ – cotθ = 0
⇒ tanθ = cotθ
⇒ tanθ = tan (90° – θ)
⇒ θ = 90° – θ
⇒ 2θ = 90°
⇒ θ = 45°

∴{\tan (\θ+15°)}/{\tan (\θ-15°)}

= {\tan (\45°+15°)}/{\tan ( 45°-15°)}

= {\tan 60°}/{\tan 30°}

= {√{3}}/{ {1}/{√{3}}}=√{3} ×{3}=3

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