Solve the given trigonometry problem

[chwala]

Homework Statement

If $x=\sec θ + \tan θ$, then show that $\dfrac {1}{x}=\sec θ - \tan θ$

Relevant Equations

Trigonometry

My take;

$x^2=\dfrac{(1+\sin θ)^2}{cos^2θ}=\dfrac{(1+\sin θ)^2}{1-\sin ^2θ}=\dfrac{1+\sin θ}{1-\sin θ}$

we know that, $x=\dfrac{1+\sin θ}{\cos θ}$

$⇒1+\sin θ=x\cos θ$

therefore,

$x^2=\dfrac{x\cos θ}{1-\sin θ}$

$\dfrac{x}{x^2}=\dfrac{1-\sin θ}{\cos θ}$

$\dfrac{1}{x}=\dfrac{1}{\cos θ}-\dfrac{\sin θ}{\cos θ}=\sec θ-\tan θ$

there may be another approach to this problem. Your input highly appreciated...

 

[BvU]

How about $x{1\over x} = 1$ and comparing that with $\sec^2 \theta -\tan^2 \theta = 1$ ?

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[chwala]

How about $x{1\over x} = 1$ and comparing that with $\sec^2 \theta -\tan^2 \theta = 1$ ?

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That is fine, comparing may mean eliminating $x$ from the equation and end up showing that the lhs=rhs in reference to the trig. identities; but that is not what we want. We need $x$ to be part of the problem.

 

[BvU]

But it is ! If you need it spelled out:$$(\sec + \tan)(\sec-\tan) = x(\sec-tan)=1 \ \& \ x{1\over x}=1 \Rightarrow (\sec-\tan)={1\over x}\\ $$

 

[chwala]

Ok @BvU ....I will check. Thanks mate.

 

[SammyS]

Ok @BvU ....I will check. Thanks mate.

Or ...
A variation, inspired by @BvU :

$\displaystyle x= \sec\theta + \tan\theta$

$\displaystyle x(\sec\theta - \tan\theta)= (\sec\theta + \tan\theta)(\sec\theta - \tan\theta)$

$\displaystyle \quad \quad \quad \quad \quad \quad \quad =\sec^2\theta - \tan^2\theta$

$\displaystyle \quad \quad \quad \quad \quad \quad \quad = 1$

Thus: $\displaystyle \quad \sec\theta - \tan\theta = \dfrac 1 x$

 

[marthasimons2]

Thanks answer
use for my ticher

 

[neilparker62]

or $$ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta = 1 \implies \sec \theta - \tan \theta = \frac{1}{\sec \theta + \tan \theta} $$