Question

Prove that :

sin A (1+tan A ) + cosA (1+cot A) =secA + cosecA .​

Answer 1

Answer:

sin A( 1+ tan A) + cos A ( 1 + cot A) = sec A +cosec A

LHS

multiplying them we get

sin A + sin A . tan A + cos A + cos A . cot A

tan A = sin A / cos A and cot A = cos A / sin A so putting the values we get

sin A + sin A . sin A / cos A + cos A + cos A . cos A / sin A

sin A + sin2A / cos A + cos A + cos2A / sin A

sin A + cos2A / sin A + cos A + sin2A / cos A

1/ sin A(sin2A + cos 2A )+ 1/ cosA ( cos2A + sin2A)

putting the value of cos2A + sin2A = 1 we get

1 / sin A + 1/ cos A

i.e. cosec A+ sec A = RHS

LHS = RHS

Answer 2

Answer:

Step-by-step explanation:

SOLVING L.H.S.

⇒ sin A (1 + $\frac{sin A}{cos A}$) + cos A (1 + $\frac{cos A}{sin A}$)

⇒ sin A ($\frac{cos A + sin A}{cos A}$) + cos A ($\frac{sin A + cos A}{sin A}$)

⇒ $\frac{sin A.cos A + sin^{2}A }{cos A}$ + $\frac{sin A.cos A + cos^{2}A }{sin A}$

⇒ $\frac{sin^{2}A.cos A + sin^{3}A + sin A.cos^{2}A + cos^{3}A }{sin A.cos A}$

⇒ $\frac{sin^{2}A(cos A+sin A)+cos^{2}A(sin A+cos A) }{sin A.cos A}$

⇒ $\frac{(sin^{2}A+cos^{2}A)(sin A+cos A) }{sinA.cosA}$

⇒ $\frac{sinA+cosA}{sinA.cosA}$

SOLVING R.H.S.

⇒ $\frac{1}{cosA} + \frac{1}{sinA}$

⇒ $\frac{sinA+cosA}{sinA.cosA}$

∴ L.H.S.=R.H.S.

HENCE PROVED!!!

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