Verify the identify.

$\frac{{\mathrm{tan}}^{2}\theta}{\mathrm{sec}\theta +1}=\frac{1-\mathrm{cos}x}{\mathrm{cos}x}$

coexpennan
2021-09-12
Answered

Verify the identify.

$\frac{{\mathrm{tan}}^{2}\theta}{\mathrm{sec}\theta +1}=\frac{1-\mathrm{cos}x}{\mathrm{cos}x}$

You can still ask an expert for help

Laith Petty

Answered 2021-09-13
Author has **103** answers

Solution:

$LHS=\frac{{\mathrm{tan}}^{2}\theta}{\mathrm{sec}\theta +1}$

$=\frac{{\left(\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}\right)}^{2}}{\frac{1}{\mathrm{cos}\theta}+1}$

$=\frac{\frac{{\mathrm{sin}}^{2}\theta}{{\mathrm{cos}}^{2}\theta}}{\frac{1+\mathrm{cos}\theta}{\mathrm{cos}\theta}}$

$=\frac{{\mathrm{sin}}^{2}\theta}{\mathrm{cos}\theta (1+\mathrm{cos}\theta )}$

$=\frac{(1-{\mathrm{cos}}^{2}\theta )}{\mathrm{cos}\theta (1+\mathrm{cos}\theta )}$

$=\frac{(1-\mathrm{cos}\theta )(1+\mathrm{cos}\theta )}{\mathrm{cos}\theta (1+\mathrm{cos}\theta )}$

$=\frac{1-\mathrm{cos}\theta}{\mathrm{cos}\theta}$

$=RHS$

Jeffrey Jordon

Answered 2022-01-31
Author has **2581** answers

Answer is given below (on video)

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Trigonometry equation simplification

So, I've been given the trig equation:

$${\mathrm{csc}}^{2}x-\mathrm{cot}2x=1$$

Using the identity and rearranging

$${\mathrm{cot}}^{2}2x+1={\mathrm{csc}}^{2}2x$$

the equation simplifies to

$${\mathrm{cot}}^{2}2x-\mathrm{cot}2x=0$$

factoring out $\mathrm{cot}2x$

$$\mathrm{cot}2x(\mathrm{cot}2x-1)=0$$

I've bee told to solve $\mathrm{\forall}x|0\le x\le 180$

So...

$$\mathrm{cot}2x=0$$

or

$$\mathrm{cot}2x=1$$

$$\frac{1}{\mathrm{tan}2x}=0\Rightarrow 1=0\times \mathrm{tan}2x$$

therefore no solutions.

$$\mathrm{cot}2x=1\Rightarrow \mathrm{tan}2x=1$$

$$x\in \{22.5,112.5\}$$

Both of my solutions were correct, but I lost marks since apparently there are two extra solutions! 45 & 135 degrees. I don't know where I went wrong in my workings and was wondering if someone could tell me what I did wrong.

So, I've been given the trig equation:

$${\mathrm{csc}}^{2}x-\mathrm{cot}2x=1$$

Using the identity and rearranging

$${\mathrm{cot}}^{2}2x+1={\mathrm{csc}}^{2}2x$$

the equation simplifies to

$${\mathrm{cot}}^{2}2x-\mathrm{cot}2x=0$$

factoring out $\mathrm{cot}2x$

$$\mathrm{cot}2x(\mathrm{cot}2x-1)=0$$

I've bee told to solve $\mathrm{\forall}x|0\le x\le 180$

So...

$$\mathrm{cot}2x=0$$

or

$$\mathrm{cot}2x=1$$

$$\frac{1}{\mathrm{tan}2x}=0\Rightarrow 1=0\times \mathrm{tan}2x$$

therefore no solutions.

$$\mathrm{cot}2x=1\Rightarrow \mathrm{tan}2x=1$$

$$x\in \{22.5,112.5\}$$

Both of my solutions were correct, but I lost marks since apparently there are two extra solutions! 45 & 135 degrees. I don't know where I went wrong in my workings and was wondering if someone could tell me what I did wrong.