(sec theta+tan theta) (1-sin thetaA)=cos theta
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I am not writing thetha
(sec+tan)(1-sin)=Cos
(1/cos + sin/cos)(1-sin)=Cos
(1+sin/cos)(1-sin)=Cos
1-sinsquarethetha/cos = cos
Cossquarethetha/costheta= cos
Lhs=RhS
if any dobt comment down else mark me as branliest
(sec+tan)(1-sin)=Cos
(1/cos + sin/cos)(1-sin)=Cos
(1+sin/cos)(1-sin)=Cos
1-sinsquarethetha/cos = cos
Cossquarethetha/costheta= cos
Lhs=RhS
if any dobt comment down else mark me as branliest
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