Question

prove that sintita -costita +1 / sintita +costita -1=1/sectita+tantita​

Answer 1

To prove.

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

Step-by-step explanation.

Given.

Here, L.H.S. is $\dfrac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}$

Step 1.

In this step, we take the L.H.S. and divide its numerator and denominator by $cos\theta\neq 0$ in order to bring terms $sec\theta$ and $tan\theta$ of the R.H.S.

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

Step 2.

In this step, we simply rearrange the terms of Step 1 in a suitable manner.

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

Step 3.

In this step, we multiply both the numerator and the denominator by $tan\theta-(1-sec\theta)$.

$\Rightarrow \dfrac{\{tan\theta-(1-sec\theta)\}^{2}}{\{tan\theta+(1-sec\theta)\}\{tan\theta-(1-sec\theta)\}}$

Step 4.

Use the algebraic identities $(a-b)^{2}=a^{2}-2ab+b^{2}$ and $(a+b)(a-b)=a^{2}-b^{2}$.

$\Rightarrow \dfrac{tan^{2}\theta-2\:tan\theta\:(1-sec\theta)+(1-sec\theta)^{2}}{tan^{2}\theta-(1-sec\theta)^{2}}$

$\Rightarrow \dfrac{tan^{2}\theta-2\:tan\theta+2\:sec\theta\:tan\theta+1-2\:sec\theta+sec^{2}\theta}{tan^{2}\theta-1+2\:sec\theta-sec^{2}\theta}$

Step 5.

In this step, we rearrange the terms in a manner so that we can use the trigonometric identity $sec^{2}\theta-tan^{2}\theta=1$.

$\Rightarrow \dfrac{sec^{2}\theta+1+tan^{2}\theta+2\:sec\theta\:tan\theta-2\:sec\theta-2\:tan\theta}{-1-(sec^{2}\theta-tan^{2}\theta)+2\:sec\theta}$

$\Rightarrow \dfrac{sec^{2}\theta+sec^{2}\theta+2\:sec\theta\:tan\theta-2\:sec\theta-2\:tan\theta}{-1-1+2\:sec\theta}$

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

Step 6.

In this step, we divide both numerator and denominator of Step 5 and try to factorize the terms with one factor $sec\theta-1$.

$\Rightarrow \dfrac{sec^{2}\theta+sec\theta\:tan\theta-sec\theta-tan\theta}{-1+sec\theta}$

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

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

Step 7.

In this step, we divide both numerator and denominator by $sec\theta-1\neq 0$ and then we use trigonometric identity $sec^{2}\theta-tan^{2}\theta=1$ and algebraic identity $a^{2}-b^{2}=(a+b)(a-b)$.

$\Rightarrow sec\theta+tan\theta$

$\Rightarrow \dfrac{sec\theta+tan\theta}{1}$

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

$\Rightarrow \dfrac{sec\theta+tan\theta}{(sec\theta+tan\theta)\:(sec\theta-tan\theta)}$

Step 8.

To reach our the conclusion to our proof, we divide both numerator and denominator by $sec\theta+tan\theta\neq 0$.

$\Rightarrow \dfrac{1}{sec\theta-tan\theta}$

Step 9. Conclusion step

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

This completes our proof.