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Sin x + cos x = cos x/1-tanx + sin x/1-cot x. Verify the identity. Explain each step please!

Sin x + cos x = cos x/1-tanx + sin x/1-cot x. Verify the identity. Explain each step-example-1
User Eular
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Answer:


sinx+cosx=(cosx)/(1-tanx)+(sinx)/(1-cotx)\\ proved.

Explanation:


sinx+cosx=(cosx)/(1-tanx)+(sinx)/(1-cotx)\\

Taking R.H.S


(cosx)/(1-tanx)+(sinx)/(1-cotx)\\

Multiply and divide first term by cos x and second term by sinx


=(cosx*cosx)/(cosx(1-tanx))+(sinx*sinx)/(sinx(1-cotx))

we know tanx = sinx/cosx and cotx = cosx/sinx


=(cos^2x)/(cosx(1-(sinx)/(cosx) ))+(sin^2x)/(sinx(1-(cosx)/(sinx)))\\=(cos^2x)/(cosx-sinx)+(sin^2x)/(sinx-cosx)

Taking minus(-) sign common from second term


=(cos^2x)/(cosx-sinx)-(sin^2x)/(cosx-sinx)

taking LCM of cosx-sinx and cosx-sinx is cosx-sinx


=(cos^2x-sin^2x)/(cosx-sinx)

We know a^2-b^2 = (a+b)(a-b), Applying this formula:


=((cosx+sinx)(cosx-sinx))/(cosx-sinx)\\=cosx+sinx\\=L.H.S

Hence proved

User Tmcallaghan
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