Question

If secA+tanA=4 find sinA.cosA

Answer 1

Answer:

Sec A -tan A =4..(i)

Also, sec ^2 A-tan ^2 A=1. (identity)

=>(sec A+tan A)(sec A-tan A)= 1

=>(sec A+tan A)=1/4..(ii)

Adding eq (I) and (ii)

2sec A=(4+1/4)=17/4

=>sec A =17/8

=>Cos A =1/sec A=8/17.

Hope it helps!!!!!!!!!!!!

Answer 2

Answer:

$\bold\red{sinA.cosA = \frac{120}{289}}$

Step-by-step explanation:

Given that,

secA + tanA = 4 ..........(i)

To find : sinA.cosA

We know the identity,

${sec}^{2}A-{tan}^{2}A=1$

also,

${a}^{2}-{b}^{2}=(a+b)(a-b)$

So, according to this,

we get,

=> (secA + tanA)(secA - tanA) = 1

But, from eqn (i),

(secA + tanA) = 4

Putting this value, we get

=> secA - tanA = 1/4 ............(ii)

Now, adding eqn (i) amd (ii),

we get,

2secA = 4+ 1/4

=> 2 secA = 17/4

=> sec A = 17/8

But, we know that, secA = 1/cosA

=> cosA = 1/secA = 8/17

Also,

we know that,

$sinA = \sqrt{1-{cos}^{2}A}$

So, putting the value of cosA ,

we get,

$sinA = \sqrt{1-{(\frac{8}{17})}^{2}}$

$=>sinA = \sqrt{1-\frac{64}{289}}$

$=>sinA = \sqrt{\frac{289-64}{289}}$

$=>sinA = \sqrt{\frac{225}{289}}$

Therefore, we get

$\bold{sinA = \frac{15}{17}}$

and

$\bold{cosA = \frac{8}{17}}$

Thus,

$=>sinA.cosA = \frac{15×8}{17×17}$

Hence,

$\bold{sinA.cosA = \frac{120}{289}}$