\operatorname { sin } ( A + B ) = 1 \text { and } \operatorname { cos

1.	\operatorname { sin } ( A + B ) = 1 \text { and } \operatorname { cos } ( A - B ) = 1,0 ^ { \circ } \leq ( A + B ) \leq 90 ^ { \circ } \text { and } A > B
Answer» Sin(a+b)=1 ——-[1]. Cos(a-b)=1———-[2] From 1 &2 Sin(a+b)=cos (a-b) Sin (a+b)= sin (90-a+b) Sin gets cancelled on both sides a+b=90-a+b 2a=90 a=45 since a and b are complementary to each other a+ b=90 B=90-45=45

\operatorname { sin } ( A + B ) = 1 \text { and } \operatorname { cos } ( A - B ) = 1,0 ^ { \circ } \leq ( A + B ) \leq 90 ^ { \circ } \text { and } A > B

Answer»

Sin(a+b)=1 ——-[1].

Cos(a-b)=1———-[2]

From 1 &2

Sin(a+b)=cos (a-b)

Sin (a+b)= sin (90-a+b)

Sin gets cancelled on both sides

a+b=90-a+b

2a=90

a=45

since a and b are complementary to each other

a+ b=90

B=90-45=45