Question

(sin a + cosec a)2 + (cos a + sec a)2 =7+tan2 a + cot 2 a

Answer 1

Correct question :

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

Solution :

Taking LHS,

• ( a + b )² = a² + b² + 2ab

=> Sin²A + Cosec²A + 2SinACosecA + Cos²A + Sec²A + 2SecACosA

• 1/SinA = CosecA
• 1/CosA = SecA

=> Sin²A + Cos²A + 2$\cancel{SinACosecA}$ + Cosec²A + Sec²A + 2$\cancel{SecACosA}$

• Sin²A + Cos²A = 1
• Cosec²A = 1 + Cot²A
• Sec²A = 1 + Tan²A

=> 1 + 2 + 1 + Cot²A + 1 + Tan²A + 2

=> 1 + 2 + 1 + 1 + 2 + Cot²A + Tan²A

=> 7 + Tan²A + Cot²A

=> RHS

Hence proved!

Answer 2

$\huge{\text{\underline{Correct question:-}}}$

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

$\huge{\text{\underline{Solution:-}}}$

★ Taking LHS,

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

$\implies$ Sin²A + Cosec²A + 2SinACosecA + Cos²A + Sec²A + 2SecACosA

$\implies$1 / SinA = CosecA

$\implies$1 / CosA = SecA

$\implies$ Sin²A + Cos²A + $2\cancel{SinACosecA}$ + Cosec²A + Sec²A +$2\cancel{SecACosA}$

$\implies$Sin²A + Cos²A = 1

$\implies$Cosec²A = 1 + Cot²A

$\implies$Sec²A = 1 + Tan²A

$\implies$ 1 + 2 + 1 + Cot²A + 1 + Tan²A + 2

$\implies$ 1 + 2 + 1 + 1 + 2 + Cot²A + Tan²A

$\implies$ 7 + Tan²A + Cot²A

$\implies$ RHS

Hence proved!

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