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Complementary Ratios

sin A + cos e...

Question

${(sin A + cos e c A)}^{2} + {(cos A + sec A)}^{2} = {tan}^{2} A + {cot}^{2} A + 7$

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Solution

$L H S = S i n^{2} A + C o s e c^{2} A + 2 S i n A C o s e c A + C o s^{2} A + S e c^{2} A + 2 C o s A S e c A$

$= (S i n^{2} A + C o s^{2} A) + 2 S i n A (\frac{1}{S i n A}) + 2 C o s A (\frac{1}{C o s A}) + C o s e c^{2} A + S e c^{2} A$
$= 1 + 2 + 2 + (1 + C o t^{2} A) + (1 + T a n^{2} A) = T a n^{2} A + C o t^{2} A + 7 = R H S$

Hence, Proved.

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Q. Solve that:-

{(sin A + cos e c A)}^{2} + {(cos A + sec A)}^{2} = 7 + {tan}^{2} A + {cot}^{2} A

(sin A + cosec A)^{2} + (cos A + sec A)^{2}

= 7 + {tan}^{2} A + {cot}^{2} A

(cos A + sec A)^{2} + {(sin A + cosec A)}^{2} - {tan}^{2} A - {cot}^{2} A .

(sin A + csc A)^{2} + (cos A + sec A)^{2} = 7 + {tan}^{2} A + {cot}^{2} A

.

(sin A + csc A)^{2} + (cos A + sec A)^{2} = 7 + {tan}^{2} A + {cot}^{2} A

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