Question

Prove that tan (45degree+A/2)=root1+sinA/1-sinA=secA+tanA

☠Wrong Ans = 50 Ans reported.☠​

Answer 1

Answer:

$= \tt \blue{ \sec \: A + \tan \: A }\: \: \:$

Step-by-step explanation:

Question:

Prove that tan $\left({{\mathrm{45°}}^{}\mathrm{{+}}\frac{A}{2}}\right)$ = $\sqrt{\frac{{1}\mathrm{{+}}\sin\hspace{0.33em}{A}}{{1}\mathrm{{-}}\sin\hspace{0.33em}{A}}}$ = sec A + tan A.

Solution:

tan $\left({{\mathrm{45°}}^{}\mathrm{{+}}\frac{A}{2}}\right)$ = $\frac{\tan{\mathrm{45°}}\mathrm{{+}}\hspace{0.33em}\tan\frac{A}{2}}{{1}\mathrm{{-}}\tan\hspace{0.33em}{\mathrm{45°}}\hspace{0.33em}\tan\frac{A}{2}}$

$= \frac{1 + \tan \frac{A}{2} }{1 - \tan \frac{A}{2} }$

$= \: \frac{ \cos \frac{A}{2} }{ \cos \frac{A}{2} - \sin\frac{ A}{2} }$

= $\sqrt{\frac{{\left({\cos\frac{A}{2}\mathrm{{+}}\sin\frac{A}{2}}\right)}^{2}}{{\left({\cos\frac{A}{2}\mathrm{{-}}\sin\frac{A}{2}}\right)}^{2}}}$

= $\sqrt{\frac{{\cos}^{2}\frac{A}{2}\mathrm{{+}}{\sin}^{2}\frac{A}{2}\mathrm{{+}}{2}\sin\frac{A}{2}\cos\frac{A}{2}}{{\cos}^{2}\frac{A}{2}\mathrm{{+}}{\sin}^{2}\frac{A}{2}\mathrm{{-}}{2}\sin\frac{A}{2}\cos\frac{A}{2}}}$

$= \sqrt{ \frac{1 + \sin \: A}{1 - \sin \:A } } \: \tt \red{ [Since, \sin2 θ = 2 \sinθ \cosθ \: ]}$

$= \sqrt{ \frac{1 + 2 \sin \frac{A}{2} \cos \frac{A}{2} }{ 1 - 2 \sin \frac{A}{2} \cos \frac{A}{2} } } \: \tt \red{[Since, \sin ^{2}θ + { \cos}^{2} θ = 1]}$

$= \sqrt{ \frac{1 + \sin \: A}{1 - \sin \: A } } \times \sqrt{ \frac{1 + \sin \: A}{1 + \sin \: A} }$

$= \frac{1 + \sin \: A}{ \sqrt{1 - \sin ^{2}A } }$

$= \frac{1 + \sin \: A}{ \cos \: A}$

$= \frac{1}{ \cos \: A } + \frac{ \sin \: A}{ \cos \: A}$

$= \tt \blue{ \sec \: A + \tan \: A }\: \: \: \red{ Ans.}$