The value of sin square 20 degree + sin square 70 degree minus 10 square 45 degree is
Answers
Answer:
0
Step-by-step explanation:
sin square 20 + sin square 70 - tan square 45
=(sin square 20 + cos square 20) - tan square 45
= 1- tan square 45
=1-1
=0
The value of \left(\sin ^{2} 20^{\circ}+\sin ^{2} 70^{\circ}-\tan ^{2} 45^{\circ}\right)(sin
2
20
∘
+sin
2
70
∘
−tan
2
45
∘
) is 0
To find the value of \sin ^{2} 20^{\circ}+\sin ^{2}(70)^{\circ}-\tan ^{2} 45^{\circ} \rightarrow(1)sin
2
20
∘
+sin
2
(70)
∘
−tan
2
45
∘
→(1)
Finding the value of each term and substituting in equation (1)
\sin ^{2} 70^{\circ}sin
2
70
∘
By using the formula of allied angles
\sin (90-\theta)=\cos \thetasin(90−θ)=cosθ
Find the number which gives 70 while subtracting with 90
\sin ^{2} 70^{\circ}=\sin ^{2}(90-20)^{\circ}sin
2
70
∘
=sin
2
(90−20)
∘
Applying the formula \sin ^{2}(90-20)^{\circ}=\cos ^{2} 20^{\circ} \rightarrow(2)\tan ^{2} 45^{\circ}sin
2
(90−20)
∘
=cos
2
20
∘
→(2)tan
2
45
∘
From the exact values of trigonometric angles the value of \tan 45^{\circ}=1tan45
∘
=1
Therefore, the value of \tan ^{2} 45^{\circ}=\left(1^{2}\right)=1 \rightarrow(3)tan
2
45
∘
=(1
2
)=1→(3)
Substituting equation (2) and (3) in equation (1)
\begin{lgathered}\begin{array}{l}{\sin ^{2} 20^{\circ}+\sin ^{2}(70)^{\circ}-\tan ^{2} 45^{\circ}} \\ {=\sin ^{2} 20^{\circ}+\cos ^{2} 20^{\circ}-1 \rightarrow(4)}\end{array}\end{lgathered}
sin
2
20
∘
+sin
2
(70)
∘
−tan
2
45
∘
=sin
2
20
∘
+cos
2
20
∘
−1→(4)
From the trigonometric identities \sin ^{2} \theta+\cos ^{2} \theta=1sin
2
θ+cos
2
θ=1
Therefore \sin ^{2} 20^{\circ}+\cos ^{2} 20^{\circ}=1sin
2
20
∘
+cos
2
20
∘
=1 using the above identity
Substituting in equation (4)
= 1-1
=0.