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Question

Answers

(I) $\sin \left( {A + B} \right) = \sin C$

(II) $\sin \left( {\dfrac{{A + B}}{2}} \right) = \cos \dfrac{C}{2}$

(III) $\tan \left( {\dfrac{{A + B - C}}{2}} \right) = \cot C$

(IV) $\tan \left( {\dfrac{{A - B - C}}{2}} \right) = - \cot A$

Select the correct answer using the codes given below,

(a) I and II (b) I, II and III (c) I, II and IV (d) all of these

Answer
Verified

(I) $\sin \left( {A + B} \right) = \sin \left( {180 - C} \right)$ [ By formula of $\sin \left( {180^\circ - \theta } \right) = \sin \theta $]

$ = \sin C$

(II) $\sin \left( {\dfrac{{A + B}}{2}} \right) = \sin \left( {\dfrac{{180 - C}}{2}} \right)$

$ = \sin \left( {90 - C} \right)$ [ By formula of$\sin \left( {90^\circ - \theta } \right) = \cos \theta $]

(III) $\tan \left( {\dfrac{{A + B - C}}{2}} \right) = \tan \dfrac{{180 - C - C}}{2}$

$ = \tan \left( {90 - C} \right)$ [By formula of $\tan \left( {90^\circ - \theta } \right) = \cot \theta $]

$ = \cot C$

(iv) $\tan \left( {\dfrac{{A - B - C}}{2}} \right) = \tan \dfrac{{A - \left( {180 - A} \right)}}{2}$

${

= \tan \left( {A - 90} \right) \\

= \tan \left\{ { - \left( {90 - A} \right)} \right\} \\

= - \cot A \\

} $ [Simplifying to get cot terms ]

$\therefore $all of these i.e. option (D) are correct.

i.e., ${90^ \circ } < \theta < {180^ \circ }.$the correct option is D. Put the formulas very attentively because a single mistake even in putting signs while conversion from one trigonometric term to another can lead to an incorrect answer.

Key concept to be applied here is that sum of all angles of a triangle is \[180^\circ \]

Concepts of trigonometry related to the topic of associated angles should be very clear, that will help you to bring an accurate answer.