Question

if 7 sin²A + 3 cos²A = 4 then prove that sec A + cosec A = 2 × 2 /√3​

Answer 1

Step-by-step explanation:

Given;

7 sin²A + 3 cos²A = 4;

To prove;

sec A + cosec A = 2 + 2 /√3;

Proof: From given;

7 sin²A + 3 cos²A = 4;

4 sin²A+3 sin²A + 3 cos²A = 4;

4 sin²A+3 (sin²A + cos²A) = 4;

4 sin²A+3 (1) = 4; (sin²A + cos²A=1)

4 sin²A = 4-3;

4 sin²A = 1;

sin²A = 1/4

sin A = 1/2;

Since, Cosec A = 1/sin A

Thus, Cosec A = 1/(1/2)

= 2 (i)

Again,

7 sin²A + 3 cos²A = 4;

7 sin²A + 3 cos²A = 4;

Add 4 cos²A on both side;

7 sin²A + 3 cos²A = 4;

7 sin²A+ 3cos²A+ 4 cos²A = 4 + 4cos²A;

7 sin²A+ 7 cos²A = 4 + 4cos²A;

7 (sin²A + cos²A) = 4 + 4cos²A;

7 (1) = 4 + 4cos²A; (sin²A + cos²A=1)

4cos²A = 7-4;

cos²A = 3/4

cos A = √3/2;

Since, Sec A =1/Cos A

Thus, Sec A = 2/√3. (ii)

From equation (i) and (ii);

Cosec A + Sec A = 2 + 2/√3;

This can also be written as;

Sec A + Cosec A = 2 + 2/√3;

That's all.