Question

Question below

Answer 1

\[\frac{ \cos A - 2 \cos^3A }{ 2\sin^3A - sinA }\]

Answer 2

Prove the above identity.

Answer 3

So you have to prove that this is \(\cot A\), right?

Answer 4

yes.

Answer 5

OK, not bad.\[\frac{\cos A (1 - 2 \cos ^2 A)}{\sin A (2 \sin^2 A - 1 )}\]\[= \frac{\cos A}{\sin A} \times \frac{1-2\cos^2 A}{2 \sin^2 A - 1}\]\[= \cot A \times \frac{1 - 2 \cos ^2 A}{2 \sin^2 A - 1}\]It is very easy to prove that the fraction is 1.

Answer 6

Im stuck at the last step..

Answer 7

Hint: write \(\cos^2 A\) in terms of \(\sin^2 A\) through an identity.

Answer 8

take cos n sin common na

Answer 9

Wo toh le liye. Aur kya common lega?

Answer 10

arre u read my q wrong its raise to 3 not 2

Answer 11

\[= \cot A \times\boxed{ \frac{1 - 2 \cos ^2 A}{2 \sin^2 A - 1}} \]

Answer 12

No I haven't read it wrong. Read my answer carefully.

Answer 13

its 2 cos^3 A n 2 sin^3 A

Answer 14

firse explain kar

Answer 15

Can you write the expression like this?\[\frac{\cos A (1 - 2 \cos ^2 A)}{\sin A (2 \sin^2 A - 1 )}\]

Answer 16

yes

Answer 17

OH I GET IT

Answer 18

But the remaing part is >_>

Answer 19

So have you got it till\[\cot A \times \frac{1 - 2\cos^2 A}{2\sin^2 A - 1}\]

Answer 20

yep .now im stuck

Answer 21

Good, now if you compare \[\cot A \times \frac{1 - 2 \cos^2 A}{2 \sin^2 A - 1}\]with the RHS \(\cot A\), you will get a hint that\[\frac{1 - 2 \cos^2 A}{2 \sin^2 A - 1}=1\]But how can you prove that?

Answer 22

In 10th, you are only supposed to know three simple identities and one of those involves \(\cos^2 \theta\) and \(\sin^2 \theta\).

Answer 23

cos ^2 A + sin^2 A is 1

Answer 24

Yes, correct. Now use that identity in this form:\[\cos^2 A = 1 -\sin ^2 A\]Replace \(\cos^2 A\) in your expression with \(1 - \sin^2 A\).

Answer 25

\[1- 2(1 - \sin^2 A)\]

Answer 26

Simplify that further and ...

Answer 27

how to...
?

Answer 28

Just open the brackets etc.

Answer 29

1 - 2 - 2sin^A

Answer 30

That's a little wrong.

Answer 31

whats wrong..?

Answer 32

\[1 - 2 \color{red}+ 2 \sin^2 A \]

Answer 33

okay!
what next..?

Answer 34

\[= \cot A \frac{2\sin^2 A - 1}{2\sin^2 A - 1}\]

Answer 35

Thanks man..!!
I have 3 more ques...

Answer 36

Please try them on your own.

Answer 37

They are not trig..
n tried these first my ans not matching

Answer 38

Achcha.