Question

For any acute angle theta prove that sin2theta +cos2theta=1 and 1+cot2theta =cosec2theta

Answer 1

Answer:

We will prove the following expression with the help of Trignometric formula and Pythagorean Theorem as:

We know that in any right angled triangle with base denoted by B, Perpendicular side denoted by P and Hypotenuse of the triangle is denoted by H we have:

1)

$\sin \theta=\dfrac{P}{H},\cos \theta=\dfrac{B}{H}$

Hence,

$\sin^2 \theta=\dfrac{P^2}{H^2},\cos^2 \theta=\dfrac{B^2}{H^2}$

Hence,

$\sin^2 \theta+\cos^2 \theta=\dfrac{P^2}{H^2}+\dfrac{B^2}{H^2}\\\\\\\sin^2 \theta+\cos^2 \theta=\dfrac{P^2+B^2}{H^2}$

By the Pythagorean Theorem we have:

$P^2+B^2=H^2$

Hence,

$\sin^2 \theta+\cos^2 \theta=1$

2)

Now , we have to prove:

$1+\cot^2 \theta=\csc^2 \theta$

We know that:

$\cot \theta=\dfrac{B}{P} , \csc \theta=\dfrac{H}{P}\\\\\\\cot^2 \theta=\dfrac{B^2}{P^2} , \csc \theta=\dfrac{H^2}{P^2}$

$1+\cot^2 \theta=1+\dfrac{B^2}{P^2}\\\\1+\cot^2 \theta=\dfrac{P^2+B^2}{P^2}$

Again by Pythagorean Theorem we have:

$P^2+B^2=H^2$

$1+\cot^2 \theta=\dfrac{H^2}{P^2}$

⇒ $1+\cot^2 \theta=\csc^2 \theta$

Answer 2

Answer:

sinθ=

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