In an acute angled triangle,cot B cot C + cot | vedclass

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

Question

(C) In $	riangle ABC$,we have $A + B + C = 180^{\circ}$.$A + B = 180^{\circ} - C$.Taking $\cot$ on both sides: $\cot(A + B) = \cot(180^{\circ} - C)$.

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) In $	riangle ABC$,we have $A + B + C = 180^{\circ}$.$A + B = 180^{\circ} - C$.Taking $\cot$ on both sides: $\cot(A + B) = \cot(180^{\circ} - C)$.Using the formula $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ and $\cot(180^{\circ} - C) = -\cot C$,we get:$\frac{\cot A \cot B - 1}{\cot A + \cot B} = -\cot C$.$\cot A \cot B - 1 = -\cot C(\cot A + \cot B)$.$\cot A \cot B - 1 = -\cot C \cot A - \cot C \cot B$.Rearranging the terms,we get:$\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$.

In an acute angled triangle,$\cot B \cot C + \cot A \cot C + \cot A \cot B$ is equal to

Similar Questions

What is the order of reactivity of halogen acids with $C_2H_5OH$?

If $A$ and $B$ are two sets,then $(A - B) \cup (B - A) \cup (A \cap B)$ is equal to

Write the resonance structures of $CO_3^{2-}$ and $HCO_3^-$.

$A$ vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles subtended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$,then $(\tan \alpha, \tan \beta)$ is equal to

The product of the distinct $(2n)^{\text{th}}$ roots of $1+i\sqrt{3}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module