RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

A and B are complementary angles.
So, A + B = 90 ; A = 90 – B
sin A = sin (90 – B) = cos B
cos A = cos (90 – B) = sin B
tan A = tan (90 – B) = cot B
cot A = cot (90 – B) = tan B
Given,
sin A cos B + cos A sin B – tan A t an B +$\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{A}\mathit{–}\mathit{c}\mathit{o}{\mathit{t}}^2\mathit{B}$
=$\mathit{s}\mathit{i}{\mathit{n}}^2$ A + $\mathit{c}\mathit{o}{\mathit{s}}^2$ B – cot B tan B + $\mathit{s}\mathit{e}{\mathit{c}}^2$ A – $\mathit{t}\mathit{a}{\mathit{n}}^2$ A
= 1 – 1 + 1 = 1