Please solve and explain. (typed response appreciated; poor eyesight) Show that (tan A)° – (sin A)° =(tan A)° (sin A)° is an identity. Use the triangle below.B,ab

Answered: Show that (tan A)“ -(sin A)´ =(tan A)ʻ…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.2: Trigonometric Functions Of Angles

Problem 45E

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Please solve and explain. (typed response appreciated; poor eyesight)

Show that (tan A)° – (sin A)° =(tan A)° (sin A)° is an identity. Use the triangle below.
B,
a
b

Expert Solution

Step 1

We have to show that ${(\tan A)}^{2} - {(\sin A)}^{2} = {(\tan A)}^{2} {(\sin A)}^{2}$ is an identity by using given triangle:

Here according to angle $A$ ,

Hypotenuse, $h = c$

Perpendicular, $p = a$

And base, $b = b$

Therefore for right angle triangle we can write from Pythagoras theorem,

$\begin{array}{rcl} {(h y p o t e n u s e)}^{2} & = & {(p e r p e n d i c u l a r)}^{2} + {(b a s e)}^{2} \\ h^{2} & = & p^{2} + b^{2} \\ c^{2} & = & a^{2} + b^{2} . . . . . . (1) \end{array}$

We know that

$\begin{array}{rcl} \tan A & = & \frac{p e r p e n d i c u l a r}{b a s e} \\ = & \frac{p}{b} \\ = & \frac{a}{b} \end{array}$

and,

$\begin{array}{rcl} \sin A & = & \frac{p e r p e n d i c u l a r}{h y p o t e n u s e} \\ = & \frac{p}{h} \\ = & \frac{a}{c} \end{array}$

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