In triangle ABC,(cot A+cot B)(cot B+cot C)(co | vedclass

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

Question

(C) Given the expression $(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A)$.Using $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we have:$= \left(\frac{

Vedclass · Accepted Answer

$\operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$

Vedclass · Answer

(C) Given the expression $(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A)$.Using $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we have:$= \left(\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B}ight) \left(\frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}ight) \left(\frac{\cos C}{\sin C} + \frac{\cos A}{\sin A}ight)$$= \left(\frac{\sin(A+B)}{\sin A \sin B}ight) \left(\frac{\sin(B+C)}{\sin B \sin C}ight) \left(\frac{\sin(C+A)}{\sin C \sin A}ight)$Since $A+B+C = \pi$,we have $\sin(A+B) = \sin(\pi-C) = \sin C$,$\sin(B+C) = \sin A$,and $\sin(C+A) = \sin B$.Substituting these values:$= \left(\frac{\sin C}{\sin A \sin B}ight) \left(\frac{\sin A}{\sin B \sin C}ight) \left(\frac{\sin B}{\sin C \sin A}ight)$$= \frac{\sin A \sin B \sin C}{(\sin A \sin B \sin C)^2} = \frac{1}{\sin A \sin B \sin C}$$= \operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$.

In $\triangle ABC$,$(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A) =$

Similar Questions

The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

In $\triangle ABC, (r_2 + r_3) \cot \left(\frac{B+C}{2}\right) = $

The ratios of the sides in a triangle are $5:12:13$ and its area is $270 \text{ cm}^2$. The sides of the triangle in $\text{cm}$ are

If $ABCD$ is a cyclic quadrilateral with $AB=6, BC=4, CD=5, DA=3$ and $\angle ABC=\theta$,then $\cos \theta=$

In $\Delta ABC,$ if $(a + b + c)(a - b + c) = 3ac,$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module