prove that ( sinA + cosA + 1) (sinA - 1 + cosA) secA cosec A = 2
Answers
Answered by
1
Step-by-step explanation:
LHS
( sinA + cosA + 1) (sinA - 1 + cosA) secAcosec A
(sin^2A + 2SinAcosA + cos^2A + SinA + CosA - sinA - cosA -1)secAcosec A
(1 + 2sinAcosA -1)secAcosec A
2sinAcosAsecAcosec A
2
RHS
So
LHS = RHS
Answered by
0
(sinA + cosA +1)(sinA + cosA - 1)secAcosecA
=[(sinA+cosA)²-1]secAcosecA
=(2sinAcosA)secAcosecA
=2 sinA×1/sinA cosA×1/cosA
=2
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