How do you prove $sec^2xcscx =sec^2x+csc^2x$ ?

Question

How do you prove $sec^2xcscx =sec^2x+csc^2x$ ?

2 Answers

Impact of this question

39381 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-21T07:37:21

Check the Problem.

Explanation:

I have doubt about the validity of the ("so called") Identity.

Had it been an Identity, it would have hold good for $x=pi/4$ .

$"For "x=pi/4, "The L.H.S."=sec^2(pi/4)csc(pi/4)$

$=(sqrt2)^2sqrt2=2sqrt2$ , whereas,

$"The R.H.S.=2+2=4"$ .

$:."The L.H.S."!="The R.H.S."$

In fact, $sec^2xcsc^2x=sec^2x+csc^2x$ .

$sec^2x+csc^2x=1/cos^2x+1/sin^2x$ ,

$=((sin^2x+cos^2x))/(cos^2xsin^2x)$ ,

$=(1)-:{1/sec^2x*1/csc^2x}$ ,

$=sec^2xcsc^2x$ .

Answer 2 · 2018-04-21T07:44:42

$sec^2xcscx!=sec^2x+csc^2x,...why ?$
Please see below.

Explanation:

We know that,

$color(blue)((1)sectheta=1/costheta and csctheta=1/sintheta$

$color(red)((2)sin^2theta+cos^2theta=1$

Here,

$LHS=color(violet)(sec^2xcscx) and RHS=sec^2x+csc^2x$

We take ,

$RHS=color(blue)(sec^2x+csc^2x$

$=color(blue)(1/cos^2x+1/sin^2x...toApply(1)$

$=color(red)((sin^2x+cos^2x))/(cos^2xsin^2x)$

$=color(red)(1)/(cos^2xsin^2x)...toApply(2)$

$=1/cos^2x*1/sin^2x$

$=color(violet)(sec^2xcsc^2x$

$!=LHS$

Hence, $sec^2xcolor(red)(cscx!=)sec^2x+csc^2x$ ,

But, $sec^2xcolor(red)(csc^2x=)sec^2x+csc^2x$

Science

Math

Humanities

... and beyond

How do you prove $sec^2xcscx =sec^2x+csc^2x$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you prove sec^2xcscx =sec^2x+csc^2x sec^2xcscx =sec^2x+csc^2x ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you prove $sec^2xcscx =sec^2x+csc^2x$ ?