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):

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