Question

If (sin a + cosec a)2 + (cos a + sec a)2 = k + tan2 a + cot2 a., then the value of k is equa
a) 9
b) 7
C) 5
​

Answer 1

Given

(sinA + cosecA)² + (cosA + secA)² = k + tan²A + cot²A

To find

The value of k.

Answer

Option b) 7

$\rule{200}2$

Explanation

⇒ (sinA + cosecA)² + (cosA + secA)² = k + tan²A + cot²A

Used identity: (a + b)² = a²+ b² + 2ab

⇒ (sin²A + cosec²A + 2sinA. cosecA) + (cos²A + sec²A + 2cosA. secA) = k + tan²A + cot²A

We know that, cosecA = 1/sinA and secA = 1/cosA

⇒ (sin²A + cosec²A + 2sinA. 1/sinA) + (cos²A + sec²A + 2cosA. 1/cosA) = k + tan²A + cot²A

⇒ (sin²A + cosec²A + 2) + (cos²A + sec²A + 2) = k + tan²A + cot²A

⇒ sin²A + cos²A + cosec²A + sec²A + 2 + 2 = k + tan²A + cot²A

Also, sin²A + cos²A = 1

⇒ 1 + cosec²A + sec²A + 4 = k + tan²A + cot²A

Also, 1 + cot²A = cosec²A and 1 + tan²A = sec²A

⇒ 5 + 1 + cot²A + 1 + tan²A = k + tan²A + cot²A

⇒ 7 + cot²A - cot²A + tan²A - tan²A = k

cot²A and tan²A cancel out, we left with

⇒ 7 = k

⇒ k = 7

Answer 2

||✪✪ QUESTION ✪✪||

if (sinA + cosecA)² + (cosA + secA)² = k + tan²a + cot2 a., then the value of k is equal to :-

a) 9

b) 7

C) 5

|| ✰✰ ANSWER ✰✰ ||

Solving LHS,

→ (sinA + cosecA)² + (cosA + secA)²

Using (a+b)² = a² + b² + 2ab now,

→ sin²A + cosec²A + 2sinA*cosecA + cos²A+sec²A + 2cosA*secA

Now, using cosecA = (1/sinA) and secA = (1/cosA)

→ sin²A + cosec²A + 2sinA*(1/sinA) + cos²A+sec²A + 2cosA*(1/cosA)

→ sin²A + cosec²A + 2 + cos²A+sec²A + 2

→ 4 + (sin²+cos²A) + cosec²A + sec²A

using ⟪ (sin²A + cos²A) = 1 , cosec²A = (1 + cot²A) & sec²A = 1 + tan²A) ⟫

→ 4 + 1 + (1 + cot²A) + ( 1 + tan²A)

→ 7 + tan²A + cot²A

Comparing it with RHS , we get,

→ 7 + tan²A + cot²A = k + tan²A + cot²A

Hence, Value of K is 7.(Option B) .