Question

Please solve limit class 12th with full explanation​

Answer 1

Answer:

รợภןคภค קɭєครє ๒гคเภɭเՇ ๓คгк ๓ץ คภรฬєг

Step-by-step explanation:

ฬє ﻮเשєภ Շﻮє Շครк

๔єคг รคภןคภค

We note that :

tan A - tan B = sin( A-B ) / ( cos A. cos B ).

Hencє,

the Numerator is

= tan(a+2h) - 2tan(a+h) + tan a

= [ tan(a+2h) - tan(a+h) ] - [ tan(a+h) - tan a ]

= { sin [ (a+2h)-(a+h) ] / cos(a+2h).cos(a+h) } - { sin[ (a+h)-a ] / cos(a+h).cos a }

= ( sin h ) [ cos a - cos(a+2h) ] / { cos a. cos(a+h). cos(a+2h) }

= ( sin h ) * 2 sin h. sin(a+h) / { cos a. cos(a+h). cos(a+2h) }

= ( 2 sec a ) * [ sin(a+h) / cos(a+h) cos(a+2h) ] * ( sin² h )

Hence, the req'd lmit is

= ( 2 sec a )*lim( x -> 0) [ sin(a+h) / cos(a+h) cos(a+2h) ]* [ lim( x -> 0) ( sin h /h ) ]²

= ( 2 sec a )* [ sin a / ( cos a. cos a ) ] * ( 1 )²

= ( 2 sec a )* ( 1 / cos a )* ( sin a / cos a )

= ( 2 sec a )* ( sec a )* ( tan a )

= 2 * sec² a * tan a

P.S. :

P.S. :This limit is the First Derivative of ( tan² x ) at ( x = a ).

P.S. :This limit is the First Derivative of ( tan² x ) at ( x = a )....It is also the Second Derivatve of ( tan x ) at ( x = a ).

Answer 2

I'm really sorry...I made an error last time

I see why did you asked the question twice...you should have told that my answer was wrong....anyway

..my bad

this time it's 100percent correct

happy to help you