(C) We know that $\sin(B + C) = \sin B \cos C + \cos B \sin C$. Substituting this into the expression,we get $\text{cosec } A \sin(B + C)$. Since $A +

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(C) We know that $\sin(B + C) = \sin B \cos C + \cos B \sin C$. Substituting this into the expression,we get $\text{cosec } A \sin(B + C)$. Since $A + B + C = \pi$ in any triangle,$B + C = \pi - A$. Therefore,$\sin(B + C) = \sin(\pi - A) = \sin A$. Substituting this back,we get $\text{cosec } A \cdot \sin A = \frac{1}{\sin A} \cdot \sin A = 1$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

Easy

In $\Delta ABC,$ $\text{cosec } A(\sin B \cos C + \cos B \sin C) = $

A
$c/a$
B
$a/c$
C
$1$
D
$c/ab$

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Class 11

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

If $A, B, C$ are the angles of a $\Delta ABC$,then with usual notations,$\frac{c^{2}-a^{2}+b^{2}}{a^{2}-b^{2}+c^{2}} = $

MediumMHT CET 2020

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In $\triangle ABC$,if $\frac{1}{r_1}, \frac{1}{r_2}$ and $\frac{1}{r_3}$ are in arithmetic progression,then $r_2 : r =$

DifficultAP EAMCET 2023

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In $\Delta ABC,$ if $a = 16, b = 24$ and $c = 20,$ then $\cos \frac{B}{2} = $

Easy

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The angles $A, B$ and $C$ of a triangle $ABC$ are in $AP$. If $b: c = \sqrt{3}: \sqrt{2}$,then the angle $A$ is (in $^{\circ}$)

MediumKCET 2008

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In a $\Delta ABC$,if the sides are $a = 3, b = 5$,and $c = 4$,then $\sin \frac{B}{2} + \cos \frac{B}{2}$ is equal to

Medium

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In $\Delta ABC,$ $\text{cosec } A(\sin B \cos C + \cos B \sin C) = $

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If $A, B, C$ are the angles of a $\Delta ABC$,then with usual notations,$\frac{c^{2}-a^{2}+b^{2}}{a^{2}-b^{2}+c^{2}} = $

In $\triangle ABC$,if $\frac{1}{r_1}, \frac{1}{r_2}$ and $\frac{1}{r_3}$ are in arithmetic progression,then $r_2 : r =$

In $\Delta ABC,$ if $a = 16, b = 24$ and $c = 20,$ then $\cos \frac{B}{2} = $

The angles $A, B$ and $C$ of a triangle $ABC$ are in $AP$. If $b: c = \sqrt{3}: \sqrt{2}$,then the angle $A$ is (in $^{\circ}$)

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