Question

Prove that (sinA + cosec A)² + (cosA + secA)² = 7 + tan²A + cot²A

Answer 1

Hi ,

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We know that ,

i ) sinAcosecA = 1

ii ) cosAsecA = 1

iii ) cosec²A = 1 + cot²A

iv ) sec²A = 1 + tan²A

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LHS = ( sinA + cosecA )² + ( cos A + sec A )²

= sin²A + cosec²A +

2sinAcosecA+cos²A+sec²A + 2cosAsecA

= ( sin²A+cos²A)+cosec²A+sec²A+2+2

= 1 + ( 1 + cot²A ) + ( 1 + tan²A ) + 4

= 7 + tan²A + cot²A

= RHS

I hope this helps you.

: )

Answer 2

$\bf\huge LHS = (sinA + cosecA)^2 + (cosA + secA)^2$

$\bf\huge (sin^2 A + cosec^2 A + 2sinA cosecA ) + (cos^2 A + sec^2 A + 2 cosA SecA)$

$\bf\huge (sin^2 A + cosec^2 A + 2 sinA . \frac{1}{sinA}) + (cos^2A + sec^2 A + 2 cosA . \frac{1}{cosA})$

$\bf\huge (sin^2 A + cosec^2 A + 2) + (cos^2 A + sec^2 A + 2)$

$\bf\huge (sin^2 A + cos^2 A + 2) + (cos^2 A + sec^2 A + 2)$

$\bf\huge sin^2 A + cos^2 A + cosec^2 A + sec^2 A + 4$

$\bf\huge 1 + (1 + cot^2) + (1 + tan^2 A) + 4$

$\bf\huge 7 + tan^2 A + cot^2 A$