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If A, B, C are the interior angles of a triangle ABC, prove that tan(B+C)/2=cot(A/2) - Mathematics

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Sum

If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`

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Solution

In ∆ABC, we have

A + B + C = 180º

⇒ B + C = 180º – A

`\Rightarrow \frac{B+C}{2}=\text{ }90^\text{o}-\frac{A}{2}`

Taking tan on both sides, we get

`\Rightarrow \tan ( \frac{B+C}{2})=\tan( 90^\text{o}-\frac{A}{2})`

`\Rightarrow \tan ( \frac{B+C}{2} )=\cot \frac{A}{2}`

Concept: Trigonometric Ratios of Complementary Angles
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APPEARS IN

ICSE Class 10 Mathematics
Chapter 18 Trigonometry
Exercise | Q 31
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