Question

sinA(1+ tan A) + cos A (1 + cot A) = sec A + cosec A​

Answer 1

Hi!!

L.H.S

=>

Sin A ( 1 + Tan A ) + Cos A ( 1 + Cot A )

=>

( Sin A + Sin² A / Cos A ) + ( Cos A + Cos² A / Sin A )

=>

( Sin A + Cos A ) + ( Sin³ A + Cos³ A )/(Sin A × Cos A )

=>

( Sin A + Cos A ) + ( Sin A + Cos A )³ - 3Sin A × Cos A ( Sin A + Cos A )/ ( Sin A × Cos A )

=>

{Sin A + Cos A } { Sin A × Cos A + 1 - Sin A × Cos A }

---------------------------------------------------

( Sin A × Cos A )

=>

( Sin A + Cos A )

------------------

( Sin A × Cos A )

=>

Sec A + Cosec A = R.H.S

Hence proved.

Note:-

( a³ + b³ ) = ( a + b )³ - 3ab ( a + b )

And

Sin² A + Cos² A = 1

Answer 2

Step-by-step explanation:

Taking LHS

Sin a (1+tana)+ cosa (1+cot a )

sin a [ 1 + sin a / cos a ] + cos a [ 1+ cos a/ sin a]

sina ( cos a + sin a )/ cos a + cos a ( sin a + cos a )/ sin a

Taking cos a + sin a common

___________________________

cos a + sina { sin a / cos a + cos a / sina }

cos a + sin a { (sin ^2a + cos ^2 a )/ cos a . sin a }

[ cos ^2 a + sin ^2 a = 1]

so

cos a + sin a ( 1/ cos a . sin a)

cos a + sin a /cos a . sin a

cos a / cosa . sina + sin a / cos a . sin a

1/ sina + 1/ cos a

cosec a + sec a .

So LHS = RHS .

HENCE PROVED.

Hope it may help you ♥♥