Question

If sec theta = 1+t^2 / 2t find all other trigonometry ratios

Answer 1

Concept

Branch of mathematics which deals with angles and sides of an right angled triangle is called as trigonometry. The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec).

Given

secФ = (1 + t²)/2t

Find

All other trigonometric ratios

Solution

secФ = (1 + t²)/2t

We need to find

sinФ , cosФ, tanФ, cotФ, cosecФ

According to trigonometry
secФ = hypotenuse/base

Hypotenuse = h = 1 + t²

Base = b = 2t

Using Pythagoras Theorem for finding perpendicular side

P = √(h² - b²)

P = √( (1 + t²)² -(2t)² )

After Solving we get

Perpendicular = p = 1 - t²

Finding other trigonometric ratios

cosФ =  b/h = 2t / 1 + t²

sinФ = p/h = 1 - t² / 1 + t²

cosecФ = h/p = 1 + t² / 1 - t²

tanФ = p/b = 1 - t² / 2t

cotФ = b/p = 2t / 1 - t²

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Answer 2

Answer:

$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{1-t^2}{1+t^2}$

$\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}=\frac{2t}{1+t^2}$

$\tan \theta = \frac{\text{Perpendicular}}{\text{Bse}}=\frac{1-t^2}{2t}$

$\cot \theta = \frac{\text{Base}}{\text{Perpendicular}}=\frac{2t}{1-t^2}$

$\sec \theta = \frac{\text{Hypotenuse}}{\text{Base}}=\frac{1+t^2}{2t}$

$\csc \theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{1-t^2}{1-t^2}$

Step-by-step explanation:

Given :

Since we are given that

$\sec\theta =\frac{1+t^2}{2t}$

To Find :

Other trigonometric Values.

Solution :

Since we are given that

$\sec\theta =\frac{1+t^2}{2t}$

and we know that sec $\theta$ = $\frac{\text{Hypotenuse}}{\text{Base}}$

So by comparing it with given part we get that

Hypotenuse  = $1+t^2$          and      Base  = $2t$

Then the Perpendicular of the Right Angled Triangle

= $\sqrt{\text{(Hypotenuse)}^2-(Base)^2}}$

= $\sqrt{(1+t)^2-(2t)^2}$

= $\sqrt{1+t^4+2t^2-4t^2}$

= $\sqrt{1+t^4-2t^2$

= $\sqrt{(1-t^2)^2}$

= $1-t^2$

Hence we get that Perpendicular = $1-t^2$

Other Trigonometric Vales

$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{1-t^2}{1+t^2}$

$\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}=\frac{2t}{1+t^2}$

$\tan \theta = \frac{\text{Perpendicular}}{\text{Bse}}=\frac{1-t^2}{2t}$

$\cot \theta = \frac{\text{Base}}{\text{Perpendicular}}=\frac{2t}{1-t^2}$

$\sec \theta = \frac{\text{Hypotenuse}}{\text{Base}}=\frac{1+t^2}{2t}$

$\csc \theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{1-t^2}{1-t^2}$

Hence we have calculated all the trigonometric ratios.

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