Question

5sinθ =3 then the value of (secθ +tanθ )/(secθ -tanθ )

Answer 1

Step-by-step explanation:

Now, sin a = 3/5

cos a = √1-sin²a = √1-9/25 = √(25-9)/25

= √16/25 = 4/5

So, sec a = 5/4

and, tan a = 3/5 × 5/4 = 3/4

As,

$\frac{ \sec(a) + \tan(a) }{ \sec(a) - \tan(a) }$

$= \frac{( \sec(a ) + \tan(a) )( \sec(a) + \tan(a) ) }{( \sec(a - \tan(a) )( \sec(a) + \tan(a)) }$

$= {( \sec(a) + \tan(a)) }^{2} \div( { \sec }^{2} a - { \tan }^{2}a)$

= (sec a + tan a)² { as, sec²a - tan²a = 1}

${( \frac{5}{4} + \frac{3}{4} ) }^{2} = {( \frac{8}{4} )}^{2} = {2}^{2} = 4$