Question

# If ABC is a triangle, then which of the following statements are correct?(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

Hint: We need to have concepts of associated angles of trigonometry well known so that we can interchange one trigonometric function into other. Relations between different trigonometric identities are clear in our mind. We know, for any triangle, if A, B & C are the three angles then $\angle A + \angle B + \angle C = 180^\circ$. First of all, look at the question attentively so that you can make a mind map of how the relation given in the question can be proved & most importantly try to find relation between those angles too upon which trigonometric functions are applied.

(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.

Note: In this type of problem, one should remember all trigonometric identities and the basic relations. The terms, tan, cot, sec, cosec are negative, if they are under quadrant of ${90^ \circ }$to ${180^ \circ }$,
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.