pls answer this trigonometric identity
Attachments:
Answers
Answered by
1
LHS
=cosA (1+cosA/sinA) + sinA(1+sinA/cosA)
=cosA(sinA+cosA)/sinA)+ sinA(cosA+sinA)/cosA
=[cos^2(sinA+cosA) + sin^2A(cosA+sinA)] / sinAxcosA
=[(cosA+sinA)(cos^2A+sin^2A)] / sinAxcosA
=cosA+sinA/sinAxcosA
=cosA/sinAxcosA + sinA/sinAxcosA
=1/sinA + 1/cosA
=cosecA+secA
=RHS
HENCE PROVED
=cosA (1+cosA/sinA) + sinA(1+sinA/cosA)
=cosA(sinA+cosA)/sinA)+ sinA(cosA+sinA)/cosA
=[cos^2(sinA+cosA) + sin^2A(cosA+sinA)] / sinAxcosA
=[(cosA+sinA)(cos^2A+sin^2A)] / sinAxcosA
=cosA+sinA/sinAxcosA
=cosA/sinAxcosA + sinA/sinAxcosA
=1/sinA + 1/cosA
=cosecA+secA
=RHS
HENCE PROVED
Anonymous:
plz mark as brainliest if my answer helped u
Similar questions