Q5 of 40 Page 422

If tan θ = √3, then

Given,


tan θ = √3


⇒ tan2θ = 3


⇒ sec2θ - 1 = 3 [As tan2θ + 1 = sec2θ]


⇒ sec2θ = 4 …[1]


Also,



⇒ as tanθ = √3


Squaring both sides,



[As cot2θ + 1 = cosec2θ]


…[2]


Putting the values from [1] and [2] into given eqn



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