If sin(A + B) = cos(A + B), what is the value of tanA?
A. $$\frac{{1 - \tan B}}{{1 + \tan B}}$$
B. $$\frac{{1 + \tan B}}{{1 - \tan B}}$$
C. $$\frac{{1 + \sec B}}{{1 - \sec B}}$$
D. $$\frac{{1 - {\text{cosec}}\,B}}{{1 + {\text{cosec}}\,B}}$$
Answer: Option A
Solution (By Examveda Team)
$$\eqalign{ & \sin \left( {A + B} \right) = \cos \left( {A + B} \right) \cr & \frac{{\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right)}} = 1 \cr & \tan \left( {A + B} \right) = 1 \cr & \tan \left( {A + B} \right) = \tan {45^ \circ } \cr & A + B = {45^ \circ } \cr & A = {45^ \circ } - B \cr & \tan A = \tan {45^ \circ } - \tan B \cr & \tan A = \frac{{\tan {{45}^ \circ } - \tan B}}{{1 + \tan {{45}^ \circ }\tan B}} \cr & \tan A = \frac{{1 - \tan B}}{{1 + \tan B}} \cr} $$This Question Belongs to Arithmetic Ability >> Trigonometry
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