2
$\begingroup$

Prove the following identities: $$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x \tag i$$ $$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x \tag {ii}$$

For $(\mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1. I then tried just multiplying out the brackets and got as far as $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$

$\endgroup$
3
  • $\begingroup$ Please see math.meta.stackexchange.com/questions/5020 $\endgroup$
    – Angina Seng
    Dec 5, 2018 at 20:34
  • 1
    $\begingroup$ I've retyped your equations using mathJax. Notice you can further edit your question, as some parts of the equations were not clear, please do so if something is not as you originally intended. $\endgroup$
    – Mefitico
    Dec 5, 2018 at 20:34
  • $\begingroup$ Welcome to MathSE. Please edit the question to show your work. That helps us identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. You should not have the $\sec^2x$ term in your answer to the first question. $\endgroup$
    – N. F. Taussig
    Dec 5, 2018 at 21:05

4 Answers 4

Reset to default
2
$\begingroup$

(i) $$(\sec^2x + \tan^2x)(\csc^2x + \cot^2x)$$ => $$(\sec^2x\csc^2x + \tan^2x\csc^2x + \sec^2x\cot^2x + \tan^2x\cot^2x)$$ => $$(\sec^2x\csc^2x+ \frac{\sin^2x}{\cos^2x\sin^2x} + \frac{\cos^2x}{\cos^2x\sin^2x} + 1)$$ => $$(\sec^2x\csc^2x+ \frac{1}{\cos^2x} + \frac{1}{\sin^2x} + 1)$$ => $$(\sec^2x\csc^2x+ \frac{\sin^2x + \cos^2x}{\cos^2x\sin^2x} + 1)$$ => $$(\sec^2x\csc^2x+ \frac{1}{\cos^2x\sin^2x} + 1)$$ => $$(\sec^2x\csc^2x+ \sec^2x\csc^2x + 1)$$ => $$(1 + 2\sec^2x\csc^2x)$$

ii) $$\frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x}$$ => $$\frac{\cos x(1 - \cot x) + \sin x(1 - \tan x)}{(1 - \tan x)(1 - \cot x)}$$ => $$\frac{\frac{\cos^2x \sin x - \cos x\cos^2x + \sin^2 x\cos x - \sin x\sin^2x}{\sin x \cos x}}{(1 - \tan x)(1 - \cot x)}$$ => $$\frac{\cos^2 x(\sin x - \cos x) + \sin^2 x(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$ => $$\frac{(\sin^2 x - \cos^2 x)(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$ => $$\frac{(\sin^2 x - \cos^2x)(\cos x - \sin x)}{(\cos x - \sin x)(\sin x - \cos x)}$$ => $$\frac{(\sin^2 x - cos^2 x)}{(\sin x - \cos x)}$$ => $$\frac{(\sin x + \cos x)(\sin x - \cos x)}{(\sin x - \cos x)}$$ => $$\sin x + \cos x$$

$\endgroup$
4
  • $\begingroup$ That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $\frac{\cos x\sin x-\cos^2x}{\sin x}+\frac{\sin x\cos x-\sin^2x}{\cos x}$, rather than $\frac{\cos x\sin x-\cos^2x}{\sin x\cos x}+\frac{\sin x\cos x-\sin^2x}{\sin x\cos x}=\frac{\cos x\sin x-\cos^2x+\sin x\cos x-\sin^2x}{\sin x\cos x}$ as you have it. $\endgroup$
    – Toby Bartels
    Dec 5, 2018 at 21:34
  • $\begingroup$ Thanks! fixed the issues... $\endgroup$
    – Gopal Anantharaman
    Dec 5, 2018 at 21:56
  • $\begingroup$ Well done I lost in that! I'll take a look to your work, Bye $\endgroup$
    – user
    Dec 5, 2018 at 22:02
  • $\begingroup$ It looks good now. $\endgroup$
    – Toby Bartels
    Dec 6, 2018 at 22:00
1
$\begingroup$

$$(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)$$

$$=(2\sec^2x-1)(2\csc^2x-1)$$

$$=4\sec^2x\csc^2x-2(\sec^2x+\csc^2x)+1$$

Now use $\sec^2x+\csc^2x=\cdots=\sec^2x\csc^2x$

The second one has been solved by Taussig

$\endgroup$
1
  • $\begingroup$ Your method for this proof is definitely an improvement over mine. $\endgroup$
    – N. F. Taussig
    Dec 8, 2018 at 14:53
0
$\begingroup$

(i)
\begin{align*} (\sec^2x + \tan^2x)(\csc^2x + \cot^2x) & = (1 + \tan^2x + \tan^2x)(1 + \cot^2x + \cot^2x)\\ & = (1 + 2\tan^2x)(1 + 2\cot^2x)\\ & = 1 + 2\cot^2x + 2\tan^2x + 4\\ & = 5 + 2(\csc^2x - 1) + 2(\sec^2x - 1)\\ & = 5 + 2\csc^2x - 2 + 2\sec^2x - 2\\ & = 1 + 2\csc^2x + 2\sec^2x\\ & = 1 + 2\left(\frac{1}{\sin^2x} + \frac{1}{\cos^2x}\right)\\ & = 1 + 2\left(\frac{\cos^2x + \sin^2x}{\sin^2x\cos^2x}\right)\\ & = 1 + \frac{2}{\cos^2x\sin^2x}\\ & = 1 + 2\sec^2x\csc^2x \end{align*}

(ii) \begin{align*} \frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x} & = \frac{\cos x}{1 - \tan x} \cdot \frac{\cos x}{\cos x} + \frac{\sin x}{1 - \cot x} \cdot \frac{\sin x}{\sin x}\\ & = \frac{\cos^2x}{\cos x - \sin x} + \frac{\sin^2x}{\sin x - \cos x}\\ & = \frac{\cos^2x}{\cos x - \sin x} - \frac{\sin^2x}{\cos x - \sin x}\\ & = \frac{\cos^2x - \sin^2x}{\cos x - \sin x}\\ & = \frac{(\cos x + \sin x)(\cos x - \sin x)}{\cos x - \sin x}\\ & = \cos x + \sin x \end{align*}

$\endgroup$
0
$\begingroup$

(i) Let $C=\cos(2x), S=\sin(2x)=2\sin(x)\cos(x)$

$S^2*RHS= S^2(1+\frac{2}{(S/2)^2})= S^2+8=9-C^2$

$\begin{align}S^2*LHS &=(2+2\sin^2(x))(2+2\cos^2(x)) \cr &=(2+(1-C))(2+(1+C))\cr &=9-C^2 =S^2*RHS \end{align}$
QED

(ii) Let $t=\tan(x/2),\text{then }\sin(x)=\frac{2t}{1+t^2}\text{ , }\cos(x)=\frac{1-t^2}{1+t^2}$

$1-\tan(x)=1-\frac{2t}{1-t^2}=\frac{1-2t-t^2}{1-t^2}$
$1-\cot(x)=1-\frac{1-t^2}{2t}=\frac{1-2t-t^2}{-2t}$

$\begin{align} \frac{\cos(x)}{1-\tan(x)} + \frac{\sin(x)}{1-\cot(x)} &=\frac{(1-t^2)^2 - (2t)^2}{(1-2t-t^2)(1+t^2)} \cr &=\frac{(1-2t-t^2)(1+2t-t^2)}{(1-2t-t^2)(1+t^2)} \cr &= \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} = \sin(x) + \cos(x) \end{align}$
QED

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .