Trigonometric proof using trigonometric identies

Question

If $x=k\sec^2\theta +m\tan^2\theta$ and $y=l\sec^2\theta +n\tan^2\theta$ prove that $\frac{x-k}{k+m}=\frac{y-l}{l+n}$

Need help in solving this. Can someone please post the entire working.

Answer 1

$$x=k\sec^2\theta +m\tan^2\theta\\y=l\sec^2\theta +n\tan^2\theta\\$$ use $$\sec^2\theta=\frac{1}{cos^2\theta}=\tan^2\theta +1$$ $$x=k(\tan^2\theta +1) +m\tan^2\theta=\tan^2\theta (k+m)+k \to \frac{x-k}{k+m}=\tan^2\theta \\y=l(\tan^2\theta +1) +n\tan^2\theta =(l+n)\tan^2\theta +l \to \frac{y-l}{l+n}=\tan^2\theta \\$$ so $$\frac{x-k}{k+m}=\tan^2\theta=\frac{y-l}{l+n} $$

Answer 2

You are very very close.

$$x-k=(k+m)\tan^2\theta,$$ $$y-l=(l+n)\tan^2\theta.$$

Answer 3

$x=k\sec^2\theta +m\tan^2\theta$

$x=k(1+\tan^2\theta)+m\tan^2\theta$

$x=k+k\tan^2\theta+m\tan^2\theta$

Similarly,

$y=l\sec^2\theta +n\tan^2\theta$

$y=l(1+\tan^2\theta)+n\tan^2\theta$

$y=l+l\tan^2\theta+n\tan^2\theta$

Now left hand side $\frac{x-k}{k+m}$

= $\frac{k+k\tan^2\theta+m\tan^2\theta-k}{k+m}$

= $\frac{k\tan^2\theta+m\tan^2\theta}{k+m}$

= $\tan^2\theta \frac{(k+m)}{k+m}$

= $\tan^2\theta$

Now right hand side $\frac{y-l}{l+n}$

= $\frac{l+l\tan^2\theta+n\tan^2\theta-l}{l+n}$

= $\frac{l\tan^2\theta+n\tan^2\theta}{l+n}$

= $\tan^2\theta \frac{(l+n)}{l+n}$

= $\tan^2\theta$

Proved.

Answer 4

One of the simplest approaches is to exploit first principles.

I prefer this approach because it avoids explicit knowledge of trigonometric identities.

Let $\Delta$ be a right triangle with sides $\{h,p,b\}$ such that $h^2 = p^2 + b^2$.

Now we can say that, $$ \cos \theta = \frac{b}{h} \qquad \sin \theta = \frac{p}{h} \qquad \tan\theta = \frac{p}{b}. $$

Now it is simple to prove your result. First, note that,

$$x - k = k\underbrace {\frac{{{h^2}}}{{{b^2}}} - \frac{{{b^2}}}{{{b^2}}}}_{{p^2}} + m\frac{{{p^2}}}{{{b^2}}} = \frac{{{p^2}}}{{{b^2}}}\left( {k + m} \right).$$

In similar way,

$$y - l = l\frac{{{h^2}}}{{{b^2}}} + n\frac{{{p^2}}}{{{b^2}}} - l\frac{{{b^2}}}{{{b^2}}} = \frac{{{p^2}}}{{{b^2}}}\left( {l + n} \right).$$

Finally, $$\frac{{x - k}}{{k + m}} = \frac{{y - l}}{{l + n}} = \frac{{{p^2}}}{{{b^2}}} = \tan \theta.$$