In - ABC - a - b - c tan A -2-tan B -2 is equal toA- tan C - 2B- 2 c C - 2C- 2 a A - 2D- 2 b B - 2

The correct option is B 2c cot C/2 We have,(a+b+c) (tanA/2+tanB/2 ) =2s (tanA/2+tanB/2 ) =2s (Δ/s(s a)+Δ/s(s b) ) =2 (Δ/s a+Δ/s b ) =2Δ (s b+s a/(s a)(s b ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

In Δ ABC , a ...

Question

$I n Δ A B C, (a + b + c) (t a n \frac{A}{2} + t a n \frac{B}{2}) is equal to$

2c cot C/2

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2a cot A/2

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2b cot B/2

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tan C/2

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Solution

The correct option is A
2c cot C/2

$We have, (a + b + c) (t a n \frac{A}{2} + t a n \frac{B}{2}) = 2 s (t a n \frac{A}{2} + t a n \frac{B}{2}) = 2 s (\frac{Δ}{s (s - a)} + \frac{Δ}{s (s - b)}) = 2 (\frac{Δ}{s - a} + \frac{Δ}{s - b}) = 2 Δ (\frac{s - b + s - a}{(s - a) (s - b)}) = 2 Δ \frac{2 s - (a + b)}{(s - a) (s - b)} = 2 Δ (\frac{c}{(s - a) (s - b)}) = 2 c c o s \frac{C}{2}$

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Similar questions

Q. In a

△ A B C, (a + b + c) (t a n \frac{A}{2} + t a n \frac{B}{2})

$For Δ A B C, t a n (\frac{B + C}{2}) =$ _______.

Q. Assertion :In a

Δ

ABC,

(a + b + c) (tan \frac{A}{2} + tan \frac{B}{2}) = 2 c cot \frac{C}{2}

Reason: In a

Δ

ABC

a = b cos C + c cos B

In a $Δ$ ABC, if the sides are a = 3, b = 5 and c = 4, then sin B/2 + cos B/2 is equal to

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In ΔABC,(a+b+c)(tanA2+tanB2) is equal to

The correct option is A 2c cot C/2 We have,(a+b+c)(tanA2+tanB2)=2s (tanA2+tanB2)=2s(Δs(s−a)+Δs(s−b))=2(Δs−a+Δs−b)=2Δ(s−b+s−a(s−a)(s−b))=2Δ2s−(a+b)(s−a)(s−b)=2Δ(c(s−a)(s−b))=2c cosC2

$I n Δ A B C, (a + b + c) (t a n \frac{A}{2} + t a n \frac{B}{2}) is equal to$