Linked Questions

0
votes
3answers
187 views

Prove $\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1+\sin\theta}{\cos\theta}$ [duplicate]

Prove by writing: $$\theta = 2A$$ that: $$\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1+\sin\theta}{\cos\theta}$$ First I subbed in the $2A$ such that: $$\frac{\sin 2A-\cos 2A+1}{\...
2
votes
1answer
127 views

Prove the following trigonometric identity: $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$ [duplicate]

Prove: $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$$ My attempt: LHS= $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$$ $$=\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin ...
2
votes
3answers
39 views

Proving trigonometric identities [duplicate]

I’ve had a bit of difficulty of this question: (1+sinA+cosA)/(1-sinA+cosA)=(1+sinA)/cosA I tried to do: (SinA)^2+(CosA)^2+sinA+cosA/(SinA)^2+(CosA)^2-sinA+cosA=(1+sinA)/cosA But then I’m kind of lost....
-1
votes
1answer
38 views

Proving $\frac{\cos a - \sin a + 1}{\cos a + \sin a - 1 } = \frac{\sin a}{1-\cos a}$ [duplicate]

Can somebody help to prove that: $$\frac{\cos a - \sin a + 1}{\cos a + \sin a - 1 } = \frac{\sin a}{1-\cos a}$$
6
votes
5answers
536 views

Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$

It's required to prove that $$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$ I managed to go about out it two ways: Assume it holds: $$\frac{1+\cos x + \sin x}{1 - \...
3
votes
5answers
246 views

Proving this trig identity:$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$

$$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$$ What I've tried, $$\frac{((1+\cos\theta)+(\sin\theta))((1+\cos\theta)+(\sin\theta))}{(1+\cos\theta-\sin\...
3
votes
4answers
195 views

How do I verify $\frac{(\sin x - \cos x + 1)}{(\sin x + \cos x-1)}=\frac{\sin x + 1}{\cos x}$?

I need to verify this trigonometric identity for an assignment:$$\frac{(\sin x - \cos x + 1)}{(\sin x + \cos x-1)}=\frac{\sin x + 1}{\cos x}$$ I've tried a few different approaches, but I end up ...
1
vote
4answers
94 views

how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$

I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are $$ \sin x \cos ...
1
vote
7answers
111 views

Proving $\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$

I got this question from a paper but can't solve it and the question paper has no solutions section.How do you prove this? $$\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \...
3
votes
4answers
168 views

Simple question on trigonometry identities of sec and tan [closed]

Please, I want to know different methods to prove following identity $$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
1
vote
3answers
134 views

Trigonometry Identity Problem

Prove that: $$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{ \cos A}$$ I found this difficult for some reason. I tried subsituting tan A for sinA / cos A and sec A as 1/...
1
vote
4answers
77 views

Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$

Which one is equivalent to: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$$ 1)$\dfrac{\cos(x)}{1+\sin(x)}$ 2)$\dfrac{\cos(x)}{1-\sin(x)}$ 3)$\dfrac{1-\sin(x)}{\cos(x)}$ 4)$\dfrac{1+\sin(x)}{\...
0
votes
2answers
61 views

I am unable to prove this Trigonometric Identity

Can you please prove this identity: $\displaystyle\frac{\cot A + \csc A - 1}{\cot A- \csc A + 1} = \frac{1+ \cos A}{\sin A}$
0
votes
0answers
28 views

Trigonometric equations having + - pattern

How to solve trigonometric equations by just looking at the form... For e.g., proving $$\cos A - \sin A + \dfrac {1}{\cos A + \sin A} - 1 = \csc A + \cot A$$ First of all I know the LHS is in basic ...