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If \[\frac{ 2 \sin … - QuestionCove
OpenStudy (jiteshmeghwal9):

If \[\frac{ 2 \sin A }{ 1+\sin A + \cos A }=k\]. then \[\frac{ 1+\sin A -\cos A }{ 1+\sin A }=?\]

1 year ago
OpenStudy (jiteshmeghwal9):

@ParthKohli

1 year ago
Parth (parthkohli):

OK, nice question. Took me a while to get to it. I thought a little about \(1 + \sin A + \cos A\) and \(1 + \sin A - \cos A\) and multiplying them should give us something nice. So if you let the given expression be \(\lambda\) then we try to find \(\frac{k}{\lambda}\) and see if it gets us somewhere.\[\frac{k}{\lambda}=\frac{2\sin A (1 + \sin A)}{(1 + \sin A + \cos A)(1 + \sin A - \cos A)}\]\[= \frac{2\sin A (1 + \sin A)}{(1 + \sin A)^2 - \cos^2A}\]\[= \frac{2\sin A (1 + \sin A) }{(1+\sin A)(1 + \sin A) - (1+ \sin A )( 1 - \sin A)}\]\[=\frac{2 \sin A}{2\sin A} = 1\]Therefore,\[k = \lambda \]

1 year ago
OpenStudy (jiteshmeghwal9):

options are (1)k/2 (2)k (3)2k (4)1/k

1 year ago
Parth (parthkohli):

Or if you want a normal proof...\[k = \frac{2\sin A }{1+\sin A + \cos A} = \frac{2\sin A (1+\sin A - \cos A)}{(1+\sin A + \cos A)(1 + \sin A - \cos A)}\]\[= \frac{2\sin A(1 + \sin A - \cos A) }{(1+\sin A)^2-\cos^2 A }\]\[= \frac{2\sin A (1 + \sin A - \cos A)}{(1+\sin A)(1+ \sin A -1 + \sin A) }\]\[= \frac{1 + \sin A - \cos A}{1+\sin A}\]

1 year ago
OpenStudy (jiteshmeghwal9):

thanx 'k' is the answer

1 year ago
Parth (parthkohli):

Exactly.

1 year ago
OpenStudy (jiteshmeghwal9):

another question :- if\[asin^2 \theta+bcos^2 \theta=m\]\[bsin^2 \phi + acos^2\phi=n\]\[a \tan \theta=b \tan \phi\]then which of the following is true :-\[1. \frac{1}{n}-\frac{1}{m}=\frac{1}{a}-\frac{1}{b}\]\[2. \frac{1}{n}+\frac{1}{m}=\frac{1}{a}+\frac{1}{b}\]\[3. m^2+n^2=a^2+b^2\]\[4. m^2-n^2=a^2-b^2\]

1 year ago
Parth (parthkohli):

Sorry, I was outside. From the first equation, you can see these:\[(a-b)\sin^2 \theta =m-b\]\[(a-b)\cos^2 \theta = a-m\]\[\Rightarrow \tan^2 \theta = \frac{m-b}{a-m}\]In the second equation,\[(a -b)\sin^2\phi = a-n\]\[(a-b)\cos^2\phi = n - b\]\[\Rightarrow \tan^2\phi = \frac{a-n}{n-b}\]Now we know \(a^2 \tan^2 \theta = b^2 \tan^2\phi\).\[\Rightarrow a^2\frac{m-b}{a-m} = b^2 \frac{a-n}{n-b}\]

1 year ago
Parth (parthkohli):

Is it B?

1 year ago
OpenStudy (jiteshmeghwal9):

yes thank you.

1 year ago
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