AP ICET 2016 Paper

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Question : 128 of 200

Marks: +1, -0

(1+\tan95°)(1+\tan130°)=

$-2$
$3+2√3$
$3-2√3$
$2$

Solution:

95°+130°=180°+45°

∴\tan(95°+130°)=\tan(180°+45°)=\tan45°=1

Applying

\tan(A+B)={\tanA+\tanB}/{1-\tanA⋅\tanB},

we get

\;\;{\tan95^{∘}+\tan130^{∘}}/{1-\tan95^{∘}\tan130^{∘}}=1

\tan95^{∘}+\tan130^{∘}=1-\tan95^{∘}\tan130^{∘}

⇒\tan95^{∘}+\tan130^{∘}+\tan95^{∘}\tan130^{∘}=1

(1+\tan95^{∘})+\tan130^{∘}(1+\tan95^{∘})=1+1

(1+\tan95^{∘})(1+\tan130^{∘})=2

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