Question

Verify the identity: Use the fundamental , & sum-difference identities in the steps.
( sec y - tan y)² = 1-sin y/ 1+sin y

Answer 1

\[(a-b)^2 = a^2 + b^2 - 2ab\]
Use this and solve LHS..

Answer 2

Solving or not??

Answer 3

= sec² a + tan²a - 2 sec a tan a
= 1+ tan²a + tan²a - 2 sec a tan a
= 1+2tan²a + 2(1+tan a )(tan a)
=1+ 4tan²a + 2 tan a
.?

Answer 4

You can do it by other method easily..

\[secy = \frac{1}{cosy}\]

\[tany = \frac{siny}{cosy}\]

So,

\[(\frac{1}{cosy} - \frac{siny}{cosy})^2 = (\frac{1-siny}{cosy})^2 = \frac{(1-siny)^2}{\cos^2y}\]

Open the parenthesis of numerator and then show me..

Answer 5

See, what you have written on RHS I think there is some mistake..
Tell me clearly what we have to verify?

Answer 6

verify that the 2 equations are equal/same

Answer 7

NO..

Answer 8

\[\left( \sec \alpha - \tan \alpha \right)^{2} = \left( 1-\sin \alpha / 1+\sin \alpha \right)\]

Answer 9

Have you got it till here:
\[\frac{(1-siny)^2}{\cos^2y}\]

Answer 10

I don't understand how to get a minus sin y
what I have is :
= ( 1/ cos² y) + (sin² y/ cos ² y) - 2sec y tan y
= ( 1+sin y/ cos y)² - 2 sec y tan y
= (1+sin² y + 2sin y/ cos² y) - 2sec y tan y

what happened to the ( 2sec y tan y ) in the one you did ?

Answer 11

See tell me have you got till there??

Answer 12

\[\cos^2x = 1 - \sin^2x = (1+sinx)(1-sinx)\]


\[\frac{(1- siny)^2}{1 - \sin^2y} = \frac{ \cancel{1-siny} \times (1-siny)}{\cancel{(1-siny)} \times (1 + siny)} = \frac{1 - siny}{1 + siny}\]