(1 + sec A)÷sec A=sin squ A ÷ (1 - cos A)
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Is it 10th standard question ???
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Answered by
2
RHS = sin^2A÷ (1- cosA)
(1- cos^2A) ÷(1-cosA)
(1-cosA)(1+cosA)÷ (1-cosA )
== 1 + cosA
== 1 + 1 ÷ sec
== ( secA+1)÷secA
which is equal to LHS
PLEASE SELECT MY ANSWERS AS BRAINLIEST
(1- cos^2A) ÷(1-cosA)
(1-cosA)(1+cosA)÷ (1-cosA )
== 1 + cosA
== 1 + 1 ÷ sec
== ( secA+1)÷secA
which is equal to LHS
PLEASE SELECT MY ANSWERS AS BRAINLIEST
Answered by
3
LHS = 1+secA/secA
=1+1/cosA/1/cosA
=cosA+1
RHS= sin^2A/1-cosA
=1-cos^2A/1-cosA
=(1-cosA)(1+cosA)/1-cosA
=1+cosA
Therefore LHS=RHS
=1+1/cosA/1/cosA
=cosA+1
RHS= sin^2A/1-cosA
=1-cos^2A/1-cosA
=(1-cosA)(1+cosA)/1-cosA
=1+cosA
Therefore LHS=RHS
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