Question

From sin²x + cos²x = 1, prove other two identities
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Answer 1

`{ \tt{ { \sin}^(2) x + \cos {}^(2)x = 1 }}`

- Divide through by cos²x ;

`{ \tt{ \frac{ { \sin }^(2)x }{ \cos {}^(2) x} + \frac{ \cos {}^(2) x }{ { \cos }^(2) x} = \frac{1}{ { \cos }^(2)x } }} \\ \\ { \boxed{ \tt{ { \tan}^(2) x + 1 = { \sec }^(2) x}}}`

- Divide tgrough by sin²x;

`{ \tt{ \frac{ { \sin }^(2)x }{ { \sin}^(2)x } + \frac{ \cos { }^(2)x }{ { \sin}^(2)x } = \frac{1}{ { \sin}^(2) x} }} \\ \\ { \boxed{ \tt{1 + { \cot }^(2)x = \csc {}^(2) x}}}`

Answer 2

See below for proof.

Explanation:

Given trigonometric identity:

`\large\boxed{\sin^2x+\cos^2x =1}`

To prove the identity 1 + cot²x = cosec²x :

`\boxed{\begin{aligned}\sin^2x+\cos^2x &=1\\\textsf{Divide by $\sin^2x$}\implies (\sin^2x)/(\sin^2x)+(\cos^2x)/(\sin^2x)& =(1)/(\sin^2x)\\1+\left((\cos x)/(\sin x)\right)^2&=\left((1)/(\sin x)\right)^2\\1+(\cot x)^2&=(\text{cosec}\: x)^2\\1+\cot^2x&=\text{cosec}\: ^2x\end{aligned}}`

To prove the identity tan²x + 1 = sec²x :

`\boxed{\begin{aligned}\sin^2x+\cos^2x &=1\\\textsf{Divide by $\cos^2x$}\implies (\sin^2x)/(\cos^2x)+(\cos^2x)/(\cos^2x)& =(1)/(\cos^2x)\\\left((\sin x)/(\cos x)\right)^2+1&=\left((1)/(\cos x)\right)^2\\(\tan x)^2+1&=(\sec x)^2\\\tan^2x+1&=\sec ^2x\end{aligned}}`

Additional identities used:

`\boxed{\begin{minipage}{4 cm}\underline{Trigonometric Identities}\\\\$\tan \theta=(\sin \theta)/(\cos \theta)$\\\\$\cot \theta=(\cos \theta)/(\sin \theta)$\\\\$\csc \theta=(1)/(\sin \theta)$\\\\$\sec \theta=(1)/(\cos \theta)$\\\end{minipage}}`