SSC Sample Paper Mock Test-7 SSC CGL Tear-II Paper-1

If ABC is a triangle, then which of the following statements are correct? I. \[\sin A+B=\sin C\] II. \[\sin \left( \frac{A+B}{2} \right)=\cos \frac{C}{2}\] III. \[\tan \frac{(A+B+C)}{2}=\cot C\] IV. \[\tan \frac{(A-B-C)}{2}=-\cot A\]

A) I and II

B) I, II and III

C) I, II and IV

D) All of these

Correct Answer: C

Solution :

As, \[A+B+C=180{}^\circ \]

I. \[\sin \,\,(A+B)=\sin \,\,(180{}^\circ -C)=\sin C\]

II. \[\sin \left( \frac{A+B}{2} \right)=\sin \left( \frac{180{}^\circ -C}{2} \right)\]

\[\Rightarrow \]\[\sin \left( 90{}^\circ -\frac{C}{2} \right)=\cos \frac{C}{2}\]

III. \[=\tan \left( \frac{A+B+C}{2} \right)=\tan \frac{(180{}^\circ -C)-C}{2}\]

\[=\tan \frac{(180{}^\circ -2\,C)}{2}\]

IV. \[\tan \frac{(A-B-C)}{2}=\tan \frac{A-(B+C)}{2}\]

\[=\tan \frac{A-(180{}^\circ -A)}{2}\]

\[=\tan [-(90{}^\circ -A)]\]

\[=-\tan \,\,(90{}^\circ -A)=-\cot A\]