If A + B + C = 180^circ,then sum tan frac{A}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $A + B + C = 180^\circ$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ$.Thus,$\frac{A}{2} = 90^\circ - (\frac{B}{2} + \frac{C}{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $A + B + C = 180^\circ$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ$.Thus,$\frac{A}{2} = 90^\circ - (\frac{B}{2} + \frac{C}{2})$.Taking tangent on both sides,$	an \frac{A}{2} = 	an(90^\circ - (\frac{B}{2} + \frac{C}{2})) = \cot(\frac{B}{2} + \frac{C}{2})$.Using the identity $\cot 	heta = \frac{1}{	an 	heta}$,we get $	an \frac{A}{2} = \frac{1}{	an(\frac{B}{2} + \frac{C}{2})}$.Therefore,$	an \frac{A}{2} 	an(\frac{B}{2} + \frac{C}{2}) = 1$.Expanding the tangent sum formula: $	an \frac{A}{2} \left( \frac{	an \frac{B}{2} + 	an \frac{C}{2}}{1 - 	an \frac{B}{2} 	an \frac{C}{2}} ight) = 1$.$	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{A}{2} 	an \frac{C}{2} = 1 - 	an \frac{B}{2} 	an \frac{C}{2}$.Rearranging the terms,we get $	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{B}{2} 	an \frac{C}{2} + 	an \frac{C}{2} 	an \frac{A}{2} = 1$.

If $A + B + C = 180^\circ$,then $\sum \tan \frac{A}{2} \tan \frac{B}{2} = $

Similar Questions

Which of the following is correct?

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $ (in $/16$)

$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $

$A$ tree breaks at a certain height and the upper part touches the ground,making an angle of $60^{\circ}$ with the ground at a distance of $10 \text{ m}$ from its foot. The original height of the tree was (in $\text{m}$):

$\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module