Question

Prove that.
(Sin A + cosec A)2
+ (cos A + sec A)2
= 7 + tan2 A + cot2 A

Answer 1

$\bold{GIVEN}$

$\mathsf{(sin \: A + cosec \: A)^{2} + (cos \: A + sec \: A)^{2} = 7 + tan^{2} \: A+ cot^{2} \: A}$

On taking left hand side,

By identity we know,

$\boxed{ (a + b)^{2} = a^{2} + b^{2} + 2ab}$

so,

$\mathsf{(sin^{2} \: A + cosec^{2 }\: A + 2 \: sin \: A \: . \: cosec \: A) + (cos^{2} \: A + sec^{2 }\: A + 2 \: cos \: A \: . \: sec\: A)}$

Since,

$\boxed{cosec \: A = \dfrac{1}{sin \: A}}$

$\boxed{sec \: A = \dfrac{1}{sec \: A}}$

So ,

$\mathsf{(sin^{2} \: A + cosec^{2 }\: A + 2 \: sin \: A \times \dfrac{1}{sin \: A}) + (cos^{2} \: A + sec^{2 }\: A + 2 \: cos \: A \times \dfrac{1}{sec \: A})}$

$\mathsf{(sin^{2} \: A + cosec^{2 }\: A + 2 \: \cancel{sin \: A} \times \dfrac{1}{ \cancel{sin \: A}}) + (cos^{2} \: A + sec^{2 }\: A + 2 \: \cancel{cos \: A} \times \dfrac{1}{ \cancel{cos \: A}})}$

$\mathsf{sin^{2} \: A + cosec^{2 }\: A + 2 + cos^{2} \: A + sec^{2 }\: A + 2}$

$\mathsf{sin^{2} \: A + cos^{2 }\: A + 4+ cosec^{2} \: A + sec^{2 }\: A }$

Since,

$\boxed{sin^{2} \: A + cos^{2 } \: A = 1}$

$\mathsf{1 + 4 + cosec^{2} \: A + sec^{2 }\: A }$

$\mathsf{5 + cosec^{2} \: A + sec^{2 }\: A }$

Since,

$\boxed{cosec^{2} \: A = 1 + cot^{2} \: A }$

$\boxed{sec^{2} \: A = 1 + tan^{2} \: A}$

$\mathsf{5 + 1 + cot^{2} \: A +1 + tan^{2} }$

$\mathsf{7 + cot^{2} \: A + tan^{2} }$ = R. H. S

$\bold{HENCE \: PROVED}$

Answer 2

Solution:-

Given:-

( SinA + CosecA)² + (CosA + SecA)²

To Prove:-

( SinA + CosecA)² + (CosA + SecA)²= 7 + tan²A + Cot²A

Proof:-

( SinA + CosecA)² + (CosA + SecA)²

=> Sin²A + Cosec²A + 2. SinA. CosecA + Cos²A + Sec²A + 2. CosA . SecA

=> Sin²A + Cos²A + 2. SinA. 1/SinA + Sec²A + Cosec²A + 2. CosA. 1/CosA

=> 1 + 2 + 2 + Sec²A + Cosec²A

=> 5 + 1 + Tan²A + 1 + Cot²A

=> 7 + Tan²A + Cot²A.

Hence Proved!!

IDENTITY USED:-

Cosec²A = 1 + Cot²A

Sec²A = 1 + Tan²A.