Question

Prove that. Sin teta- costeta+1÷sinteta+costeta-1

Answer 1

Step-by-step explanation:

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Answer 2

To prove that:

$\dfrac{\sin \theta- \cos \theta+1}{\sin \theta+\cos \theta-1} =\dfrac{1}{\sec \theta-\tan \theta}$.

Solution:

L.H.S. = $\dfrac{\sin \theta- \cos \theta+1}{\sin \theta+\cos \theta-1}$

On dividing numerator and denominator by $\cos \theta$, we get

= $\dfrac{\tan \theta- 1+\sec \theta}{\tan \theta+ 1-\sec \theta}$ [ ∵ $\tan \theta=\dfrac{\sin \theta}{\cos \theta}$ and $\sec \theta =\dfrac{x}{ \cos \theta }$]

= $\dfrac{\sec \theta+\tan \theta- 1}{\tan \theta-\sec \theta+ 1}$

= $\dfrac{(\sec \theta+\tan \theta)- (\sec^2 \theta-\tan^2 \theta)}{\tan \theta-\sec \theta+ 1}$ [ ∵ 1 = $\sec^2 \theta-\tan^2 \theta$]

= $\dfrac{(\sec \theta+\tan \theta)-(\sec \theta+\tan \theta)(\sec \theta-\tan \theta) }{\tan \theta-\sec \theta+ 1}$

= $\dfrac{(\sec \theta+\tan \theta)[1-(\sec \theta-\tan \theta)]}{(\tan \theta-\sec \theta+ 1)}$

= $\dfrac{(\sec \theta+\tan \theta)(1-\sec \theta+\tan \theta)}{(\tan \theta-\sec \theta+ 1)}$

= $\dfrac{(\sec \theta+\tan \theta)(\tan \theta-\sec \theta+ 1)}{(\tan \theta-\sec \theta+ 1)}$

= $\sec \theta+\tan \theta$

R.H.S. = $\dfrac{1}{\sec \theta-\tan \theta}$

Rationalizing numerator and denominator, we get

= $\dfrac{1}{\sec \theta-\tan \theta}\times \dfrac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta}$

Using the algebraic identity:

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

$= \dfrac{\sec \theta+\tan \theta}{\sec^2 \theta-\tan^2 \theta}$

= $\sec \theta+\tan \theta$

∴ L.H.S. =R.H.S. = $\sec \theta+\tan \theta$, proved.

Thus, $\dfrac{\sin \theta- \cos \theta+1}{\sin \theta+\cos \theta-1} =\dfrac{1}{\sec \theta-\tan \theta}$, proved.