Question

If

3 (sec^2 A + tan^2 A) = 5,

then the general value of A is -​

Answer 1

Step-by-step explanation:

$3( {sec}^{2} A + {tan}^{2} A) = 5 \\ {sec}^{2} A + {tan}^{2} A = \frac{5}{3} \\ 1 + {tan}^{2} A + {tan}^{2} A = \frac{5}{3} \\ 1 + 2{tan}^{2} A = \frac{5}{3} \\ 2{tan}^{2} A = \frac{5}{3} - 1 \\ 2{tan}^{2} A = \frac{5 - 3}{3} \\ 2{tan}^{2} A = \frac{2}{3} \\ {tan}^{2} A = \frac{2}{3 \times 2}\\ {tan}^{2} A = \frac{1}{3} \\ tanA \: = \frac{1}{ \sqrt{3} } \\ tanA \: = tan \: 30 \degree \\ \therefore \: A \: = 30 \degree$

Answer 2

Answer:

3(sec 2 A+tan2 A)=5

sec2 A+tan2 A= 5/3

1+tan2 A+tan2 A= 5/3

1+2tan2 A= 5/3

2tan2 A= 5/3-1

2tan 2 A= 5−3/3

2tan2 A= 2/3

tan2 A= 2/3×2

tanA= 1/_/3

tanA=tan30°

∴A=30°