Question

tan teta +sec teta-1/tan teta-sec teta+1
=1+son teta/cos teta​

Answer 1

Answer:

$\displaystyle{\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1} =\frac{1+\sin\theta}{\cos\theta} \ \text{, Proved} }$

Step-by-step explanation:

Given :

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

We know :

1 = sec² Ф  -  tan² Ф

Replacing 1 by sec² Ф  -  tan² Ф in numerator ;

$\displaystyle{\text{L.H.S}=\frac{\tan\theta+\sec\theta-(\sec^2\theta-\tan^2\theta)}{\tan\theta-\sec\theta+1} }$

Using identity a² - b² = ( a + b ) ( a - b )

$\displaystyle{\text{L.H.S}=\frac{\tan\theta+\sec\theta-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{\tan\theta-\sec\theta+1} }\\\\\\\displaystyle{\text{L.H.S}=\frac{(\tan\theta+\sec\theta)[1-(\sec\theta-\tan\theta)]}{\tan\theta-\sec\theta+1} }\\\\\\\displaystyle{\text{L.H.S}=\frac{(\tan\theta+\sec\theta)[(\tan\theta-\sec\theta+1)]}{\tan\theta-\sec\theta+1} }\\\\\\\displaystyle{\text{L.H.S}=(\tan\theta+\sec\theta)}$

Now covert tan and sec into sin and cos term

$\displaystyle{\implies\dfrac{\sin\theta}{\cos\theta} +\dfrac{1}{\cos\theta} }\\\\\\\displaystyle{\implies\dfrac{1+\sin\theta}{\cos\theta} }$

L.H.S. = R.H.S.

Hence proved.