Question

20. (3) ( A+B=90^{circ} Rightarrow A=90^{circ}-B )
( Rightarrow sin A=sin left(90^{circ}-Bright)=cos B )
Similarly, ( Rightarrow cos A=sin B, tan A=cot B )
( therefore sin A cdot cos B+cos A cdot sin B- )
( tan A cdot tan B+sec ^{2} A-cot ^{2} B= )
( cos ^{2} B+sin ^{2} B-cot B . tan B+ )
( sec ^{2} A-tan ^{2} A=1-1+1=1 )
( left[because tan B cdot cot B=1, sec ^{2} A-tan ^{2} Aright. )
( =1} )

# 20. If A and B are complementary angles, then the value of sin A cos B + cos A sin B - tan A tan B + sec2 A - cot2 B is (1) 2 (2) O (3) 1 (4)-1

Solution