Trigonometrical Relations-10

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Prove that :

(i) 1 / (cosec A - cot A) - 1 / sin A = 1 / sin A = 1 / sin A - 1 / (cosec A + cot A) ;

(ii) (sin A + cosec A) ² + (cos A + sec A) ² = 7 + tan ² A+ cot ² A ;

(tan A + sec A - 1) / (tan A - sec A + 1) = sec A + tan A = (1 + sin A) / cos A .

Solution :

(i) L H S = 1 / (cosec A - cot A) - 1 / sin A = 1 / sin A = 1 / sin A - 1 / (cosec A + cot A)

=> 1 / (cosec A - cot A) + 1 / (cosec A + cot A) = 1 / sin A + 1 / sin A = 2 / sin A

L H S = (cosec A + cot A + cosec A - cot A) / (cosec A - cot A) (cosec A + cot A) = 2 cosec A / (cosec ² A - cot ² A) = 2 cosec A / 1 = 2 X 1 / sin A = 2 / sin A = R H S

(ii) L H S = (sin A + cosec A) ² + (cos A + sec A) ²

= sin ² A + cosec ² A + 2 sin A cosec A + cos ² A + sec ² A + 2 cos A sec A

= sin ² A + cot ² A + 1 + 2 sin A X (1 / sin A) + cos ² A + tan ² A + 1 + 2 cos A X (1 / cos A)

= sin ² A + cos ² A + cot ² A + tan ² A + 1 + 2 + 1 + 2

= 1 + cot ² A + tan ² A + 6 = 7 + tan ² A + cot ² A = R H S

(iii) L H S = (tan A + sec A - 1) / (tan A - sec A + 1) = [tan A + sec A - (sec ² A- tan ² A)] / (tan A - sec A + 1)

= [ (sec A + tan A) - (sec A + tan A) (sec A - tan A) ] / (tan A - sec A + 1)

= (sec A + tan A) (1 - sec A + tan A) / ( tan A - sec A + 1 ) = sec A + tan A = R H S

= 1 / cos A + (sin A / cos A) = (1 + sin A) / cos A = R H S