how do you prove this: secx(cscx)-2cosx(cscx)=tanx-cotx?

Question

use the basic trigonometric identities.

Answer 1

secx(cscx) - 2cosx(cscx)=tanx - cotx

Rule of thumb when solving identities:
(1) Always choose the more complex side
(2) Convert everything to sines and cosines.

We're either going to choose the left hand side (LHS) or the right hand side (RHS). Since the LHS is more complicated, we choose the LHS.

LHS = sec(x)csc(x) - 2cos(x)csc(x)

Convert everything to sines and cosines, using the definition of each trig function.

LHS = [1/cos(x)][1/sin(x)] - 2cos(x)[1/sin(x)]

Merge each fraction. After all, they're just multiplication.

LHS = 1/[sin(x)cos(x)] - 2cos(x)/sin(x)

Now, put them all under a common denominator. In this case, the common denominator would be sin(x)cos(x), and the second term is missing a cos(x).

LHS = [1 - 2cos^2(x)]/[sin(x)cos(x)]

Now, let's convert 1 into the famous identity, sin^2(x) + cos^2(x).

LHS = [sin^2(x) + cos^2(x) - 2cos^2(x)]/[sin(x)cos(x)]

And combine like terms

LHS = [sin^2(x) - cos^2(x)]/[sin(x)cos(x)]

Now, let's split this into two fractions.

LHS = [sin^2(x)]/[sin(x)cos(x)] - [cos^2(x)]/[sin(x)cos(x)]

And now we can cancel stuff in each fraction common to both numerator and denominator.

LHS = [sin(x)/cos(x)] - [cos(x)/sin(x)]

Which are defined to be

LHS = tan(x) - cot(x) = RHS

Answer 2

Cscx-cotx Secx 1

Answer 3

secx*cscx - 2cosx*cscx = tanx - cotx

Converting everything to sine and cosine:

1/(cosx*sinx) - 2cosx/sinx = sinx/cosx - cosx/sinx

Clear the denominators by multiplying everything by sinx*cosx

1 - 2cos^2(x) = sin^2(x) - cos^2(x)
1 - cos^2(x) = sin^2(x)
sin^2(x) = sin^2(x)

And the identity is proved.

Answer 4

secx(cscx)-2cosx(cscx)=ta...

Multiplying the L.H.S by cosx* sinx and using
Cosx*secx = 1
Sinx*(cscx) =1

L.H.S becomes
{1 - 2 (cosx) ^2}

Since 1 = (cosx) ^2 + (sinx) ^2

{1 - 2 (cosx) ^2} = (sinx) ^2 - (cosx) ^2.

Dividing again by cosx*sinx we get

tanx - cotx.

Answer 5

secx(cscx)-2cosx(cscx)=tanx-cotx

L.H.S= secx(cscx)-2cosx(cscx)
= cscx(secx-2cosx) taking cscx common
= cscx(secx-2/secx) cosx=1/secx
= cscx((secx^2-2)/secx taking LCM
= (cscx/secx)(secx^2-1-1)
= cotx(tanx^2-1) (cscx/secx)=cotx
and tanx^2=secx^2-1
=cotx.tanx^2-cotx
=tanx-cotx=R.H.S
L.H.S=R.H.S
hence proved

Answer 6

change tanx to sinx/cosx then change cotx to cosx/sinx change secx to 1/cosx and cscx to 1/sinx sinx/cosx + cosx/sinx=1/cosx(1/sinx) when u add u add or subtract u need to get common denominators for that side of the equation so times sinx/cosx by sinx and times cosx/sinx by cosx sin^2x +cos^2x/cosxsinx=1/cosx(1/sinx) sin^2x + cos^2x = 1 so substitute it for 1 1/cosxsinx=1/cosx(1/sinx) so now for the other side of the equation multiple across with 1/cosx(1/sinx) and u get..... 1/cosxsinx=1/cosxsinx =]

Answer 7

sec x = 1/cos x and cosec x = 1/sin x
then,
sec x . cosec x - 2cos x . cosec x = 1/cos x sin x - 2cos x / sin x

= 1 - 2cos^2 x / cos x sin x
= (sin^2 x + cos^2 x - 2cos^2 x) / cos x sin x
= (sin^2 x - cos^2 x)/ cos x sin x
= sin x / cos x - cos x / sin x
= tan x - cot x

Answer 8

secx(cscx) - 2cosx(cscx) =? tanx-cotx
(1/cosx)(1/sinx) - 2cosx)(1/sinx) =? (sinx)/(cos(x) - (cosx)/(sinx)
1/(sinxcosx) - 2cos^2x/(sinxcosx) =? (sinx)/(cos(x) - (cosx)/(sinx)
(1 - 2cos^2x)/(sinxcosx) =? (sinx)/(cos(x) - (cosx)/(sinx)
(sin^2x + cos^2x - 2cos^2x)/(sinxcosx) =? (sinx)/(cos(x) - (cosx)/(sinx)
(sin^2x - cos^2x)/(sinxcosx) =? (sinx)/(cos(x) - (cosx)/(sinx)
sin^2x/(sinxcosx) - cos^2x/(sinxcosx) =? (sinx)/(cos(x) - (cosx)/(sinx)
sinx/cosx - cosx/sinx = (sinx)/(cos(x) - (cosx)/(sinx)