NTSE Stage 2 SAT 2019 Exam Question Paper

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Question : 47 of 100

Marks: +1, -0

If

\tan^{2}θ=1-e^{2}

, then the value of

\secθ+\tan^3θ\;\cosecθ

is equal to

$(1-e^{2})^{1/2}$
$(2-e^{2})^{1/2}$
$(2-e^{2})^{3/2}$
$(1-e^{2})^{3/2}$

Solution:

∵\tan^{2}θ=1-e^{2}

⇒\secθ=√{1+\tan^{2}θ}=√{1+1-e^{2}}

⇒\secθ=√{2-e^{2}}

......(a)

∴\secθ+\tan^{3}θ\cscθ=\;\;{1}/{\cosθ}+\tan^{2}θ⋅\;\;{\sinθ}/{\cosθ}⋅\;\;{1}/{\sinθ}

=\;\;{1}/{\cosθ}(1+\tan^{2}θ)=\;\;{\sec^{2}θ}/{\cosθ}

=\sec^{3}θ=(2-e^{2})^{3/2}

[from(a)]

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