Question

Please solve this question class 10

Answer 1

please find the attachment

Answer 2

Question:

Prove that $\dfrac{1 + cosA + sinA}{1+cosA - sin A} = \dfrac{1+ sinA}{cosA}$

Solution: 1

RHS = $\dfrac{1+ sinA}{cosA}$

LHS = $\dfrac{1 + cosA + sinA}{1+cosA - sin A}$

Dividing numerator and denominator from cos A

$= \dfrac{\dfrac{1}{cosA} + \dfrac{cosA}{cosA} + \dfrac{sinA}{cosA}}{\dfrac{1}{cosA} + \dfrac{cosA}{cosA} - \dfrac{sinA}{cosA}}$

$= \dfrac{secA + 1 + tan A}{secA + 1 - tan A}$

$= \dfrac{secA + tanA + ({sec}^{2}A - {tan}^{2}A)}{secA + 1 - tan A}$

$= \dfrac{secA + tanA + (secA- tanA)(secA + tanA)}{secA + 1 - tan A}$

$= \dfrac{(secA+ tanA)(1 + secA - tanA)}{secA + 1 - tan A}$

$= \dfrac{(secA+ tanA)\cancel{(1 + secA - tanA)}}{\cancel{secA + 1 - tan A}}$

$= \dfrac{1}{cosA}+ \dfrac{sinA}{cosA}$

$= \dfrac{1 + sin A}{cosA}$ = RHS

Hence, proved.

☆Identities used☆

• $a^{2} - b^{2} = (a+b)(a-b)$

• $1 + {tan}^{2}\theta = {sec}^{2}\theta$

• $sec\theta = \dfrac{1}{cos\theta}$

• $tan\theta = \dfrac{sin\theta}{cos\theta}$