Question

If sin 19° = a, find the value of cos 71°, tan 71° cos 19° in terms of a.

Answer 1

Step-by-step explanation:

given that, sin19°=a

cos71°= cos(90-19)°

= sin19°

= a

tan71°= sin71°÷cos71°

= sin(90-19)°÷cos71°

= cos19°÷sin19°

= √1-sin^2(19)° ÷ sin19°

= √1-a^2 ÷ a

= √1-a^2/a

cos19°= √1-sin^2(19)°

= √1-a^2

Answer 2

Given: sin19°=a

To find: We have to find the value of cos 71°, tan 71° and cos 19°.

Solution:

We know that-

$sin^219°+cos^219°=1$

The value of sin19° is a. Putting the value of sin19° in the above equation we get-

$(a)^2+cos^219°=1\\cos^219°=1-a^2\\cos19°=√{1-a^2}$

Again we can write-

sin19°

=sin(90-19)°

=cos71°

The value of sin19° is a. So, the value of cos71° is a.

The value of sec71° is 1/a.

We know that-

$sec^271°-tan^271°=1\\(1/a)^2-tan^271°=1\\tan^271°=1/a^2-1\\tan^271°=(1-a^2)/a^2\\tan71°=√{1-a^2}/a$

The value of cos71° is a.

The value of tan71° is $√({1-a^2})/a$.

The value of cos19° is $√{1-a^2}$.