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Prove i) cosA/1+sinA + 1+sinA/cosA= secA ii)tan/1-cot + cot/1-tan = 1+ sec,cosec.... |
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Answer» hєчα mαtєReplace tan A by SIN A/cos A and cot A by cos A/sin A. We getReplace 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 getReplace 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 AReplace 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 AReplace 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 AProvedplz mαrk αѕ вrαnlíѕt |
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