Question

Prove i) cosA/1+sinA + 1+sinA/cosA= secA ii)tan/1-cot + cot/1-tan = 1+ sec,cosec....​

Answer 1

Answer:

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Replace tan A by sin A/cos A and cot A by cos A/sin A. We get

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)On simplifying we get

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)On simplifying we get(sin^3 A – cos^3 A)/ [sin A.cos A (sin A – cos A)]

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)On simplifying we get(sin^3 A – cos^3 A)/ [sin A.cos A (sin A – cos A)]= (sin A – cos A)( sin^2 A + cos^2 A + sin A.cos A] / [sin A.cos A (sin A – cos A)]

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)On simplifying we get(sin^3 A – cos^3 A)/ [sin A.cos A (sin A – cos A)]= (sin A – cos A)( sin^2 A + cos^2 A + sin A.cos A] / [sin A.cos A (sin A – cos A)]= (sin A – cos A)( 1 + sin A.cos A] / [sin A.cos A (sin A – cos A)]

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)On simplifying we get(sin^3 A – cos^3 A)/ [sin A.cos A (sin A – cos A)]= (sin A – cos A)( sin^2 A + cos^2 A + sin A.cos A] / [sin A.cos A (sin A – cos A)]= (sin A – cos A)( 1 + sin A.cos A] / [sin A.cos A (sin A – cos A)]=( 1 + sin A.cos A] / sin A.cos A

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)On simplifying we get(sin^3 A – cos^3 A)/ [sin A.cos A (sin A – cos A)]= (sin A – cos A)( sin^2 A + cos^2 A + sin A.cos A] / [sin A.cos A (sin A – cos A)]= (sin A – cos A)( 1 + sin A.cos A] / [sin A.cos A (sin A – cos A)]=( 1 + sin A.cos A] / sin A.cos A= 1 + sec A.cosec A

Replace tan A by sin A/cos A and cot A by cos A/sin A. We get[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].LCM of denominator is sin A.cos A (sin A – cos A)On simplifying we get(sin^3 A – cos^3 A)/ [sin A.cos A (sin A – cos A)]= (sin A – cos A)( sin^2 A + cos^2 A + sin A.cos A] / [sin A.cos A (sin A – cos A)]= (sin A – cos A)( 1 + sin A.cos A] / [sin A.cos A (sin A – cos A)]=( 1 + sin A.cos A] / sin A.cos A= 1 + sec A.cosec AProved

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Answer 2

Step-by-step explanation:

sinA+1-cosA)/(cosA-1+sinA) =(1+sinA)/cosA

L.H.S.

=[sinA+(1-cosA)]/[sinA-(1-cosA)]

Multiplying by [sinA+(1-cosA)]

=[sinA+(1-cosA)]^2/[sin^2A-(1-cosA)^2]

=[sin^2A+2.sinA.(1-cosA)+(1-cosA)^2]/[(1-cos^2A)-(1-cosA)^2]

=[(1-cos^2A)+2.sinA.(1-cosA) +(1-cosA)^2]/[(1-cosA)(1+cosA) - (1-cosA)^2].

=(1-cosA)[1+cosA+2sinA+1-cosA]/(1-cosA)[1+cosA-1+cosA].

=[2+2sinA]/[2.cosA].

=2(1+sinA)/2.cosA.

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= (1+sinA)/cosA