Question

3 sin a -4 cos a=0 , sec a =?​

Answer 1

Step-by-step explanation:

Given that 3 sina - 4 cos a =0

=>3 sin a= 4 cos a

=>sin a= 4/3 cos a

=> sin a/ cos a= 4/3

=> tan a = 4/3

we know that sec^2 a - tan^2 a =1

sec^2 a= 1+ tan^2 a

=>sec^2 a= 1+ (4/3)^2

=> sec^2 a= 1+(16/9)

=>sec^2 a = (9+16)/9

=> sec^2 a= 25/9

=> sec a=√(25/9)

=>sec a= 5/3

Answer 2

$Given\: 3 sinA - 4 cos A = 0$

$\implies 3sinA = 4 cos A$

$\implies \frac{sin A}{cos A} = \frac{4}{3}$

$\implies tan A = \frac{4}{3}\: ---(1)$

$Now, \red{ Value \: of \: Sec A } \\= \sqrt{ 1 + tan^{2} A } \\= \sqrt{ 1 + \big( \frac{4}{3}\big)^{2} } \\= \sqrt{ \frac{9 + 16}{9} } \\= \sqrt{\frac{25}{9}} \\= \sqrt{\big(\frac{5}{3}\big)^{2}} \\= \frac{5}{3}$

Therefore.,

$\red{ Value \: of \: Sec A } \green {= \frac{5}{3}}$

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