trigonometry?

Question

tan a/1- cot a+cot a/1-tan a =1+tan a+cot a=1+ sec a+ cosec a
is there any way to prove this quickly?

Answer 1

If you write the left side in
terms of sin(a) and cos(a) and combine the resulting fractions
over common denominators in the numerator and denominator, and
then simplify, you will obtain an expression which can be reduced
to the second expression 1 + tan(a) + cot(a), so the first
equality expresses an identity, not an equation to be solved.
Neither of the first two expressions is identically 0.

The third expression is not equal to the other two for all a.
For example, if a = pi/4, 1+tan(a)+cot(a) = 3, while
1 + sec(a) + COSEC(a) = 2*sqrt(2)+1 .

IF YOU NEED FURTHER EXPLANATION REGARDING THIS SUM YOU CAN GO TO

Answer 2

LHS = tan a /(1-cot a) + cot a /(1-tan a)
multiply both num and den by tan a
= tan^2 a /(tan a - 1) + 1/tan a (1- tan a)
= (tan ^3 a - 1)/((tan a -1)tan a)

= ((tan ^3 a -1 )/(tan a - 1))/ tan a

= (tan ^ 2 a + tan a + 1)/ tan a (using (x^3-y^3) formula
= tan a + 1 + cot a

= middle term

= 1 + sin a/cos a + cos a/ sin a
= 1 + (sin ^2 a + cos^2 a)/ sin a cos a
= 1 + 1/ sin a cos a
= 1 + cosec a sec a
not RHS

Answer 3

1. Solve for left side
(tan a/1- cot a)+(cot a/1-tan a)
=tan a/(1- (1/tan a)) + (cot a/1-tan a)
=tan a/((tan a - 1)/ tan a) + (cot a/1-tan a)
=tan^2 a / (tan a - 1) + (cot a/1-tan a)
= -tan^2 a/ (1 - tan a) + (cot a/1-tan a)
= (cot a - tan^2 a) / (1-tan a)
1+tan a+cot a
2. Solve for middle
1+tan a+cot a
=1+tan a+cot a * (1-tan a)/(1-tan a)
=(1-tan a + tan a - tan^2 a + cot a - tan a cot a)/(1-tan a)
=(1 -tan^2 a + cot a -1)/(1-tan a)
=(cot a - tan^2)/(1-tan a)
So... Left and middle are equal...
3. Solve for Right
1+ sec a+ cosec a
= I haven't figured it out yet...
Wow... he's good...

Answer 4

The right hand side of the question is wrong. When we convert the ratios of left hand side into sin & cos & simplify it, we get,

1+ cosecA.secA and not 1+cosecA+secA

Answer 5

in the first part you have 0. something is missing. Check to see if you typed it in right.

Answer 6

dont u think u are missing something u are missning brackets do it first and compleat ur quistion otherwise it will not be eaqual

Answer 7

Not equal so no solutions

Answer 8

Yes. There is.
But I forgot what it is, because I don't have to take math anymore.

Answer 9

tana/1-cota + cota/1-tana
=sin^2a/cosa(sina-cosa) + cos^2a/sina(cosa-sina)
=sin^3a-cos^3a/sinacosa(sina-cosa)
=sin^2a+cos^2a+sinacosa/sinacosa
=1+tana+cota

Answer 10

Sorry , they r not equal.....