Question

prove that tan teta +sec teta -1/tan teta -sec teta +1=1+sin teta/cos teta​

Answer 1

Answer:

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

$=\frac{\textstyle\sin^2\theta+(1-\cos\theta)^2+2\sin\theta(1-\cos\theta)}{\textstyle\sin^2\theta-(1-\cos\theta)^2}\\\\=\frac{\textstyle\sin^2\theta+1+\cos^2\theta-2\cos\theta+2\sin\theta(1-\cos\theta)}{\textstyle\sin^2\theta-1-\cos^2\theta+2\cos\theta}\\\\=\frac{\textstyle2-2\cos\theta+2\sin\theta(1-\cos\theta)}{\textstyle-2\cos^2\theta+2\cos\theta}$

$=\frac{\textstyle2(1-\cos\theta)+2\sin\theta(1-\cos\theta)}{\textstyle2\cos\theta(1-\cos\theta)}\\\\=\frac{\textstyle2(1+\sin\theta)(1-\cos\theta)}{\textstyle2\cos\theta(1-\cos\theta)}\\\\=\frac{\textstyle1+\sin\theta}{\textstyle\cos\theta}$

Hope that helps.