without using trigonometric table
evaluate : (cos^2 20+cos^2 70/sec^2 50-cot^2 40) ×2sec^2 60
Answers
Step-by-step explanation:
\dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60=8
sec
2
50−cot
2
40
cos
2
20+cos
2
70
×2sec
2
60=8
Step-by-step explanation:
We have,
\dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60
sec
2
50−cot
2
40
cos
2
20+cos
2
70
×2sec
2
60
To find, \dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60=?
sec
2
50−cot
2
40
cos
2
20+cos
2
70
×2sec
2
60=?
∴ \dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60
sec
2
50−cot
2
40
cos
2
20+cos
2
70
×2sec
2
60
=\dfrac{\cos^2 20+\cos^2 (90-20)}{\sec^2 50-\cot^2 (90-50)}\times 2 sec^2 60
sec
2
50−cot
2
(90−50)
cos
2
20+cos
2
(90−20)
×2sec
2
60
=\dfrac{\cos^2 20+\sin^2 20}{\sec^2 50-\tan^2 50}\times 2 sec^2 60=
sec
2
50−tan
2
50
cos
2
20+sin
2
20
×2sec
2
60
Using trigonometric identity,
\cos (90-A)=\sin Acos(90−A)=sinA and
\cot (90-A)=\tan Acot(90−A)=tanA
=\dfrac{1}{1}\times 2 sec^2 60=
1
1
×2sec
2
60
Using trigonometric identity,
\cos^2 A+\sin^2 A=1cos
2
A+sin
2
A=1 and
\sec^2 A-\tan^2 A=1sec
2
A−tan
2
A=1
= 2\times (2)^2=2×(2)
2
= 8
Hence, \dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60=8
sec
2
50−cot
2
40
cos
2
20+cos
2
70
×2sec
2
60=8