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Question
Prove the following identity :
( sin A + sec A )2 + ( cosA + cosecA )2 = ( 1 + secA • cosecA )
( sin A + sec A )2 + ( cosA + cosecA )2 = ( 1 + secA • cosecA )
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Solution
(sinA + secA)² + (cosA + cosecA)²
= (sinA + 1/cosA)² + (cosA + 1/sinA)²
= sin²A + 2sinA/cosA + 1/cos²A + cos²A + 2cosA/sinA + 1/sin²A
= (sin²A + cos²A) + (2sinA/cosA + 2cosA/sinA) + (1/cos²A + 1/sin²A)
= 1 + 2(sin²A+cos²A)/(sinAcosA) + (sin²A+cos²A)/(sin²Acos²A)
= 1 + 2/(sinAcosA) + 1/(sin²Acos²A)
= (1 + 1/(sinAcosA))²
= (1 + secA cosecA)²
You forgot to put the square on the RHS
= (sinA + 1/cosA)² + (cosA + 1/sinA)²
= sin²A + 2sinA/cosA + 1/cos²A + cos²A + 2cosA/sinA + 1/sin²A
= (sin²A + cos²A) + (2sinA/cosA + 2cosA/sinA) + (1/cos²A + 1/sin²A)
= 1 + 2(sin²A+cos²A)/(sinAcosA) + (sin²A+cos²A)/(sin²Acos²A)
= 1 + 2/(sinAcosA) + 1/(sin²Acos²A)
= (1 + 1/(sinAcosA))²
= (1 + secA cosecA)²
You forgot to put the square on the RHS
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