please can someone solve this quickly? I am preparing for my 10th board exam so please help.
(cosA - sinA)(1 + tanA) / 2cos²A - 1 = secA
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Step-by-step explanation:
L.H.S=(cosA - sinA)(1 + tanA) / 2cos²A - 1
=(cosA- sinA)(1+sinA/cosA)/2Cos^2A-1
[because tanA=sinA/Cos A]
cosA(1+SinA/cosA)-sinA(1+sinA/cosA)/2Cos^2A-1
multiplying inside the brackets,
cosA+SinA-sinA-sin^2A/cosA/2Cos^2A-1
=cosA-sin^2A/cosA/2Cos^2A-1
=cos^2A-sin^2A/cosA/2Cos^2A-1
=cos^2A-(1-Cos^2A)/cosA/2cos^2A-1
[because sin^2A+cos^2A=1 so sin^2A=1-Cos^2A
=cos^2A-1+Cos^2A/cosA/2Cos^2A-1
=2Cos^2A-1/cosA(2Cos^2A-1)
2Cos^2A gets cancelled
=1/CosA=SecA=R.H.S
i am not sure you understand.....u could write down and try it like this.
I hope u undrstand and plss markbrainliest if usefull.
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