Find the value of sec^2 20-cot^2 70÷sin^2 70+sin^2 20.
Answers
Answer:
sin20
o
sin40
o
sin60
o
sin80
o
=sin(60
o
−20
o
)sin20
o
sin(60
o
+20
o
)sin60
o
=
4
1
sin(3×20
o
)sin60
o
=
4
1
sin
2
60
o
=
16
3
(∵sin(60−θ)sin(60+θ)sinθ=
4
1
sin3θ)
k=3
Answer:
= 1
Step-by-step explanation:
We have,
\dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}
sin
2
70+sin
2
20
sec
2
20−cot
2
70
To find, the value of \dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}
sin
2
70+sin
2
20
sec
2
20−cot
2
70
= ?
∴ \dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}
sin
2
70+sin
2
20
sec
2
20−cot
2
70
= \dfrac{\sec^2 20-\cot^2 (90-20)}{\sin^2 70+\sin^2 (90-70)}
sin
2
70+sin
2
(90−70)
sec
2
20−cot
2
(90−20)
Using the trigonometric identity,
\tan AtanA = \cot (90-A)cot(90−A) and
\cos AcosA = \sin (90-A)sin(90−A)
= \dfrac{\sec^2 20-\tan^2 20}{\sin^2 70+\cos^2 70}
sin
2
70+cos
2
70
sec
2
20−tan
2
20
Using the trigonometric identity,
\sec^2 Asec
2
A - \tan^2 Atan
2
A = 1 and
\sin^2 Asin
2
A + \cos^2 Acos
2
A = 1
= \dfrac{1}{1}
1
1
= 1
∴ \dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}
sin
2
70+sin
2
20
sec
2
20−cot
2
70
= 1
Thus, the value of \dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}
sin
2
70+sin
2
20
sec
2
20−cot
2
70
is equal to "one (1)".