Question

sintheta+1-costheta/costheta-1+sintheta=1+sintheta/costheta
​

Answer 1

GIVEN :

The equation is [tex]\frac{sin\theta+1-cos\theta}{cos\theta-1+sin\theta}=\frac{1+sin\theta}{cos\theta }[/tex]

TO FIND :

The given equation [tex]\frac{sin\theta+1-cos\theta}{cos\theta-1+sin\theta}=\frac{1+sin\theta}{cos\theta }[/tex] is true and check the equality.

SOLUTION :

Given equation is [tex]\frac{sin\theta+1-cos\theta}{cos\theta-1+sin\theta}=\frac{1+sin\theta}{cos\theta }[/tex]

Now taking the LHS

$\frac{sin\theta+1-cos\theta}{cos\theta-1+sin\theta}$

$=\frac{sin\theta+1-cos\theta}{sin\theta-1+cos\theta}$

$=\frac{sin\theta+(1-cos\theta)}{sin\theta-(1-cos\theta)}$

Multiply and divide by the denominator's conjugate $sin\theta+(1-cos\theta)$

$=\frac{sin\theta+(1-cos\theta)}{sin\theta-(1-cos\theta)}\times \frac{sin\theta+(1-cos\theta)}{sin\theta+(1-cos\theta)}$

By using the algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{[sin\theta+(1-cos\theta)]^2}{sin^2\theta-(1-cos\theta)^2}$

​By using the algebraic identity :

$(a+b)^2=a^2+2ab+b^2$

$=\frac{sin^2\theta+2(sin\theta)(1-cos\theta)+(1-cos\theta)^2}{sin^2\theta-(1-cos\theta)^2}$

$\frac{sin^2\theta+2(sin\theta)(1-cos\theta)+(1^2-2(1)(cos\theta)+(cos\theta)^2)]}{sin^2\theta-(1^2-2(1)(cos\theta)+(cos\theta)^2)}$

​By using the algebraic identity :

$(a-b)^2=a^2-2ab+b^2$

$=\frac{sin^2\theta+2sin\theta(1)-2sin\theta(cos\theta)+1-2cos\theta+cos^2\theta}{sin^2\theta-(1-2cos\theta+cos^2\theta)}$

$=\frac{1+2sin\theta-2sin\theta cos\theta+1-2cos\theta}{sin^2\theta-((sin^2\theta+cos^2\theta)-2cos\theta+cos^2\theta)}$

By using the Trignometric identity :

$sin^2x+cos^2x=1$

$=\frac{1+2sin\theta-2sin\theta cos\theta+1-2cos\theta}{sin^2\theta-sin^2\theta-cos^2\theta+2cos\theta-cos^2\theta)}$

$=\frac{2+2sin\theta-2sin\theta cos\theta-2cos\theta}{-2cos^2\theta+2cos\theta}$

$=\frac{2(1+sin\theta-sin\theta cos\theta-cos\theta)}{2(-cos^2\theta+cos\theta)}$

$=\frac{1+sin\theta-sin\theta cos\theta-cos\theta}{cos\theta-cos^2\theta}$

$=\frac{1+sin\theta-sin\theta cos\theta-cos\theta}{cos\theta(1-cos\theta)}$

$=\frac{(1-cos\theta)+sin\theta(1-cos\theta)}{cos\theta(1-cos\theta)}$

$=\frac{(1-cos\theta)(1+sin\theta}{cos\theta(1-cos\theta)}$

$=\frac{1+sin\theta}{cos\theta}$ = RHS

⇒ LHS = RHS

⇒ [tex]\frac{sin\theta+1-cos\theta}{cos\theta-1+sin\theta}=\frac{1+sin\theta}{cos\theta }[/tex]

∴ [tex]\frac{sin\theta+1-cos\theta}{cos\theta-1+sin\theta}=\frac{1+sin\theta}{cos\theta }[/tex] is true and the equality is verified.