Question

1-sin theta divided by 1+sin theta= (sec theta-tan theta) whole square​

Answer 1

Answer:

Solution in the above image...

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Answer 2

Question :

Prove that :

$\:\:\:\:\:\:\;\:\:\:\red\bigstar \large \sf \dfrac{1-sin\theta}{1+sin\theta} =( sec\theta - tan \theta )^2$

Solution :

Solving L.H.S.

$\large \implies \sf \dfrac{1-sin\theta}{1+sin\theta}$

• To Solve this fraction we have to Rationalise.

• To Rationalise this fraction we have Multiply and divide this fraction by conjugate of its denominator.

So,

Denominator is 1 + sinθ and its conjugate will be 1-sinθ.

Now,

$\large \implies \sf \displaystyle \dfrac{1-sin\theta}{1+sin\theta} \times \dfrac{1-sin\theta}{1-sin\theta}$

$\red\bigstar \boxed{(a+b)(a-b)= a^2-b^2 }$

Applying this Identity here,

$\large \implies \sf \dfrac{(1-sin\theta)^2}{1^2 -sin^2\theta}$

we know that ,

• Sin²θ + cos²θ = 1

so,

• cos²θ = 1-sin²θ

Using this Identity, we got :

$\large \implies \sf (\dfrac{(1-sin\theta)}{cos \theta} )^2$

• Splitting

$\large \implies \sf (\dfrac{1}{cos \theta}- \dfrac{sin \theta}{cos \theta} )^2$

we know ,

• Secθ = 1/cosθ
• tanθ = sinθ/cosθ

So,

$\sf \large \implies ( sec\theta - tan \theta )^2$

$\huge \underline{Hence \: \: proved }$

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