Question

1+sinθ-cosθ/1+sinθ+cosθ=tanθ/2

Answer 1

why are thetas being displayed as ???

Answer 2

\[1+\sin \alpha-\cos \alpha/1+\sin \alpha+\cos \alpha=\tan \alpha/2\]

Answer 3

Please give brackets and write the question, is it to prove lhs =rhs ??

Answer 4

I have got his question.. :)

Answer 5

\[\frac{1 + \sin(x) - \cos(x)}{1 + \sin(x) + \cos(x)} = \tan(\frac{x}{2})\]

Answer 6

it may help to make
u = 1+sin
and the half angle identity for tangent is given as
(1-cos)/sin

\[\rightarrow \frac{u-\cos x}{u+ \cos x} = \frac{1- \cos x}{\sin x}\]
multiply left side by conjugate
expand and simplify until both sides are equal

Answer 7

Multiply by its conjugate after combining the terms as:
\[\frac{(1 + \sin(x) - \cos(x)) \times ( 1 - (\sin(x) + \cos(x))}{(1 + (\sin(x) + \cos(x))) \times (1 - (\sin(x) + \cos(x)))}\]

Answer 8

Sorry, not conjugate, it is just like rationalizing the denominator..

Answer 9

The denominator simplifies to \(2 \sin(x) \cos(x)\) or simply \( \sin(2x)\)..

Answer 10

And in numerator when you will multiply the two big brackets, they will simplify to :
\[2 \cos(x) - 2 \cos^2(x)\]

Answer 11

Thereby giving you:
\[\implies \frac{2 \cos(x) - 2 \cos^2(x)}{2 \sin(x) \cos(x)} \implies \frac{1 - \cos(x)}{\sin(x)}\]

Answer 12

you could also try \[\begin{align}\frac{1 + \sin(x) - \cos(x)}{1 + \sin(x) + \cos(x)} &= \dfrac{2\sin^2(x/2)+2\sin(x/2)\cos(x/2)}{2\cos^2(x/2) + 2\sin(x/2)\cos(x/2)} \\~\\
&= \dfrac{\sin(x/2)[2\sin(x/2) + 2\cos(x/2)]}{\cos(x/2)[2\cos(x/2) + 2\sin(x/2)]}\\~\\
&=\dfrac{\sin(x/2)}{\cos(x/2)}

\end{align}\]