Proof...trig function?

Question

(1 + sin theta) /cos theta = cos theta/ 1-sin theta

Where do I start ?

Answer 1

(1 + sin theta) /cos theta = cos theta/ 1-sin theta

bring up the denominator by cross multiplying......

(1 + sin theta)*(1-sin theta) = cos theta*cos theta

1 - sin^2 theta = cos^2 theta

cos^2 theta + sin^2 theta = 1 ............. (proven)

Answer 2

Permit my use of x instead of theta, now begin at the left hand side and you can build up to the right hand side. All you need here is some elementary algebra and the Pythagorean relation cos^2 x + sin^2 x = 1
Let's begin,
the left side is-
(1 + sin x) / cos x
multiply the numerator and the denominator by cos x, thereby effectively not altering the fraction(as they'd cut out and leave 1,but don't do that!)
then,
(1+ sin x)cos x / cos^2 x
Now recall that cos^2 x = 1- sin^2 x = (1+sin x)(1-sin x)
using the difference of two squares,
so, (1+ sin x) cos x / (1+sin x)(1-sin x)
is now our expression, the (1+sin x) is common on the top and bottom so get rid of it! And we have,
cos x / (1-sin x)
the required right hand side! Voila! You're done!
Hope this helps!!

Answer 3

Start with cos^2 theta = 1 - sin^2 theta = (1 - sintheta)(1 + sintheta) [the usual factorization of (1 - x^2) = (1 - x )(1 + x )]


Then write cos theta x cos theta = (1 - sintheta) (1 + sintheta) Divide bothe sides of the equation by costheta x (1 - sin theta) et voilá!

Answer 4

multiply and divide by (1-sin theta)
we get
=(1+sin theta)(1-sin theta)/[(cos theta)(1-sin theta)]
=(1-sin^2 theta)/[(cos theta)(1-sin theta)]
=(cos^2 theta)/[(cos theta)(1-sin theta)]
=(cos theta)/(1-sin theta)

formulas used:
1-sin^2 theta=cos^2 theta.

Answer 5

(1+sin θ)/cos θ = (cos θ)/(1-sin θ)

RHS:
(cos θ)/(1-sin θ)
Rationalise the fraction by multiplying with (1+sin θ) for both the numerator and denominator
= (cos θ)/(1-sin θ) x (1+sin θ)/(1+sin θ)
= [cosθ (1+sin θ)]/(1-sin² θ)
= [cosθ (1+sin θ)]/cos² θ
Cancel off cos θ and you will get your LHS
= (1+sin θ)/cos θ


Note that
sin² θ + cos² θ =1
So, 1-sin² θ = cos² θ
and similarly 1-cos² θ=sin² θ

Hope this helps.

Proving questions comes with lots of practice and a certain level of intuition or guessing as to which approach you should take. But with sufficient practice, it should not pose much of a problem.

Answer 6

I'll work on the RHS and make it equal to the LHS:

First note that \frac{1+sinθ}{1+sinθ}=1

Multiply the equation by 1 (as above):
\frac{cosθ*(1+sinθ)}{(1-sinθ)*(1+sinθ)}

Now expand the bottom, but leave the top alone:
\frac{cosθ*(1+sinθ)}{1+sinθ-sinθ-sin^2θ}

Simplify:
\frac{cosθ*(1+sinθ)}{1-sin^2θ}

Use the identity 1=sin^2θ+cos^2θ:
\frac{cosθ*(1+sinθ)}{cos^2θ}

Cancel any terms:
\frac{1+sinθ}{cosθ}

Now it's equal to the LHS.

Answer 7

multiply numerator and denominator by 1-sinQ. u get cos^2Q in numerator, rest is easy

Answer 8

i dont know.... i just know that i hate trig...sry