for any acute angle theta ,prove that sin² theta +cos²=1, 1+cot² theta=cosec² theta
Answers
Answer:
sinθ=
H
P
,cosθ=
H
B
inθ=
H
P
,cosθ=
H
B
Hence,
\sin^2 \theta=\dfrac{P^2}{H^2},\cos^2 \theta=\dfrac{B^2}{H^2}sin
2
θ=
H
2
P
2
,cos
2
θ=
H
2
B
2
Hence,
\begin{gathered}\sin^2 \theta+\cos^2 \theta=\dfrac{P^2}{H^2}+\dfrac{B^2}{H^2}\\\\\\\sin^2 \theta+\cos^2 \theta=\dfrac{P^2+B^2}{H^2}\end{gathered}
sin
2
θ+cos
2
θ=
H
2
P
2
+
H
2
B
2
sin
2
θ+cos
2
θ=
H
2
P
2
+B
2
By the Pythagorean Theorem we have:
P^2+B^2=H^2P
2
+B
2
=H
2
Hence,
\sin^2 \theta+\cos^2 \theta=1sin
2
θ+cos
2
θ=1
2)
Now , we have to prove:
1+\cot^2 \theta=\csc^2 \theta1+cot
2
θ=csc
2
θ
We know that:
\begin{gathered}\cot \theta=\dfrac{B}{P} , \csc \theta=\dfrac{H}{P}\\\\\\\cot^2 \theta=\dfrac{B^2}{P^2} , \csc \theta=\dfrac{H^2}{P^2}\end{gathered}
cotθ=
P
B
,cscθ=
P
H
cot
2
θ=
P
2
B
2
,cscθ=
P
2
H
2
\begin{gathered}1+\cot^2 \theta=1+\dfrac{B^2}{P^2}\\\\1+\cot^2 \theta=\dfrac{P^2+B^2}{P^2}\end{gathered}
1+cot
2
θ=1+
P
2
B
2
1+cot
2
θ=
P
2
P
2
+B
2
Again by Pythagorean Theorem we have:
P^2+B^2=H^2P
2
+B
2
=H
2
1+\cot^2 \theta=\dfrac{H^2}{P^2}1+cot
2
θ=
P
2
H
2
⇒ 1+\cot^2 \theta=\csc^2 \theta1+cot
2
θ=csc
2
θ
Hence,
\sin^2 \theta=\dfrac{P^2}{H^2},\cos^2 \theta=\dfrac{B^2}{H^2}sin
2
θ=
H
2
P
2
,cos
2
θ=
H
2
B
2
Hence,
\begin{gathered}\sin^2 \theta+\cos^2 \theta=\dfrac{P^2}{H^2}+\dfrac{B^2}{H^2}\\\\\\\sin^2 \theta+\cos^2 \theta=\dfrac{P^2+B^2}{H^2}\end{gathered}
sin
2
θ+cos
2
θ=
H
2
P
2
+
H
2
B
2
sin
2
θ+cos
2
θ=
H
2
P
2
+B
2
By the Pythagorean Theorem we have:
P^2+B^2=H^2P
2
+B
2
=H
2
Hence,
\sin^2 \theta+\cos^2 \theta=1sin
2
θ+cos
2
θ=1
2)
Now , we have to prove:
1+\cot^2 \theta=\csc^2 \theta1+cot
2
θ=csc
2
θ
We know that:
\begin{gathered}\cot \theta=\dfrac{B}{P} , \csc \theta=\dfrac{H}{P}\\\\\\\cot^2 \theta=\dfrac{B^2}{P^2} , \csc \theta=\dfrac{H^2}{P^2}\end{gathered}
cotθ=
P
B
,cscθ=
P
H
cot
2
θ=
P
2
B
2
,cscθ=
P
2
H
2
\begin{gathered}1+\cot^2 \theta=1+\dfrac{B^2}{P^2}\\\\1+\cot^2 \theta=\dfrac{P^2+B^2}{P^2}\end{gathered}
1+cot
2
θ=1+
P
2
B
2
1+cot
2
θ=
P
2
P
2
+B
2
Again by Pythagorean Theorem we have:
P^2+B^2=H^2P
2
+B
2
=H
2
1+\cot^2 \theta=\dfrac{H^2}{P^2}1+cot 2 θ= P 2H 2⇒ 1+\cot^2 \theta=\csc^2 \theta1+cot
2
θ=csc
2
θ
Answer:
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