How do I verify the identity: sec(t)+1/tan(t)=tan(t)/sec(t)-1

Question

17-04-2017
Mathematics

contestada

How do I verify the identity: sec(t)+1/tan(t)=tan(t)/sec(t)-1

Respuesta :

Otras preguntas

In order for two triangles to be similar, all corresponding pairs of angles must be __________. adjacent congruent proportional supplementary

The male elephant at the city zoo weighs 8,000 pounds, which is the same as 4 tons. The female elephant weighs 7,000 pounds. How many tons does she weigh

First drop down menu: 8.5, 40, 4, 7.5, 0 Second drop down menu: 16,6,4,0,5 Third drop down menu: 0,5,1,2.5 4th drop down menu:-4,0,2,4

Read the excerpt from “Take the Tortillas Out of Your Poetry.” In other cases, the censoring has been direct and brutal. On February 28, 1981 the morning newspa

find the missing length a)12 b)25 c)9 d)64

How did the Brown v. Board of Education Supreme Court decision overrule the Plessy v. Ferguson case? Question 1 options: separate but equal facilities were ofte

Suppose that 16 of the 80 state governors in a country were women. On each day of 2013, one of these state governors was randomly selected to read the invocati

What contributed to the economic success of Great Zimbabwe?its valuable copper mines(wrong)its location on the Nile Riverits location as a port cityits commerci

In the latte 1800’s the growth of the railroad system led to midwestern cities such as Minneapolis and Cleveland becoming ?

Under the statutes of many states, a bar owner is criminally liable when an employee bartender sells liqurd to juveniles. What if the bar owner instructs teh ba

jdoe0001 jdoe0001 · Answer 1 · 2017-04-17T00:38:42+02:00

[tex]\bf \textit{Pythagorean Identities} \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1 \\ \quad \\ 1+cot^2(\theta)=csc^2(\theta) \\ \quad \\ \boxed{1+tan^2(\theta)=sec^2(\theta)}\implies tan^2(\theta)=sec^2(\theta)-1\\\\ -----------------------------\\\\ \cfrac{sec(x)+1}{tan(x)}=\cfrac{tan(x)}{sec(x)-1}\\\\ -----------------------------\\\\[/tex]

[tex]\bf \cfrac{sec(x)+1}{tan(x)}\cdot \cfrac{sec(x)-1}{sec(x)-1}\impliedby \textit{using the conjugate}\\\\ -----------------------------\\\\ recall\qquad \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -----------------------------\\\\ thus\qquad \cfrac{sec^2(x)-1^2}{tan(x)sec(x)-1}\implies \cfrac{sec^2(x)-1}{tan(x)sec(x)-1} \\\\\\ \cfrac{\underline{tan^2(x)}}{\underline{tan(x)} sec(x)-1}\implies \cfrac{tan(x)}{sec(x)-1}[/tex]