This is not a graded question. There are 4 steps. 2:41 ) & N I A A OQ * * 4E 53%illeenisd.schoology.com3May 5: Practice 61 of 5Complete the steps by applying trig ldentities to show that the left hand side of the equationequivalent to the right hand side.1+twr - tan? z1j cot'Steps:1+tan" r= tan": Recz1seel: tan z- cot2: - sin r -: sec r.1+cul t-cot::sin a: sin' æ: -i: . cot?: se::NexII+

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Description: This is not a graded question. There are 4 steps. 2:41 ) & N I A A OQ * * 4E 53%illeenisd.schoology.com3May 5: Practice 61 of 5Complete the steps by applying trig ldentities to show that the left hand side of the equationequivalent to the right hand

We have to prove the identity 1+tan2x1+cot2x=tan2x. Use formula tanx=sinxcosx and cotx=cosxsinx Take…

Answered: Complete the steps by applying trig…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.3: Zeros Of Polynomials

Problem 67E

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This is not a graded question. There are 4 steps.

2:41 ) & N I A A O
Q * * 4E 53%
illeenisd.schoology.com
3
May 5: Practice 6
1 of 5
Complete the steps by applying trig ldentities to show that the left hand side of the equation
equivalent to the right hand side.
1+twr - tan? z
1j cot'
Steps:
1+tan" r
= tan"
: Recz1seel
: tan z- cot2
: - sin r -
: sec r.
1+cul t-cot
::
sin a
: sin' æ
: -i
: . cot?
: se
::
Nex
II
+

Expert Solution

Step 1

We have to prove the identity $\frac{1 + \tan^{2} x}{1 + \cot^{2} x} = \tan^{2} x$ .

Use formula $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$

Take left hand side

$\begin{matrix} \frac{1 + \tan^{2} x}{1 + \cot^{2} x} = \frac{1 + {(\frac{\sin x}{\cos x})}^{2}}{1 + {(\frac{\cos x}{\sin x})}^{2}} \\ = \frac{1 + \frac{\sin^{2} x}{\cos^{2} x}}{1 + \frac{\cos^{2} x}{\sin^{2} x}} \\ = \frac{\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}}{\frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x}} \\ = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x} \times \frac{\sin^{2} x}{\sin^{2} x + \cos^{2} x} (Use property \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \times \frac{d}{c}) \\ = \frac{\sin^{2} x}{\cos^{2} x} \\ = \tan^{2} x \\ = R H S \end{matrix}$

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