Question :
- tan alpha + cot Alpha = cosec alpha. sec alpha
Required solution :
Here we would be considering L.H.S. and solving.
Identities as we know that,
- [tex] \red{\boxed{\sf{tan \: \alpha \: = \: \dfrac{sin \: \alpha }{cos \: \alpha} }}}[/tex]
- [tex] \red{\boxed{\sf{cot \: \alpha \: = \: \dfrac{cos \: \alpha }{sin \: \alpha} }}}[/tex]
By using the identities we gets,
[tex] : \: \implies \: \sf{ \dfrac{sin \: \alpha }{cos \: \alpha} \: + \: \dfrac{cos \: \alpha }{sin \: \alpha} }[/tex]
[tex]: \: \implies \: \sf{ \dfrac{sin \: \alpha \times sin \: \alpha }{cos \: \alpha \times sin \: \alpha} \: + \: \dfrac{cos \: \alpha \times cos \: \alpha }{sin \: \alpha \times \: cos \: \alpha } } [/tex]
[tex] : \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha }{cos \: \alpha \times sin \alpha} \: + \: \dfrac{cos {}^{2} \: \alpha }{sin \: \alpha \times \: cos \: \alpha } } [/tex]
[tex]: \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha }{cos \: \alpha \: sin \alpha} \: + \: \dfrac{cos {}^{2} \: \alpha }{sin \: \alpha \: cos \: \alpha } } [/tex]
[tex]: \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha \: + \: cos {}^{2} \alpha}{cos \: \alpha \: sin \alpha} } [/tex]
Now, here we would be using the identity of square relations.
- [tex]\red{\boxed{ \sf{sin {}^{2} \alpha \: + \: cos {}^{2} \alpha \: = \: 1}}}[/tex]
By using the identity we gets,
[tex] : \: \implies \: \sf{ \dfrac{1}{cos \: \alpha \: sin \alpha} }[/tex]
[tex]: \: \implies \: \sf{ \dfrac{1}{cos \: \alpha } \: + \: \dfrac{1}{sin\: \alpha} }[/tex]
[tex]: \: \implies \: \bf{sec \alpha \: cosec \: \alpha}[/tex]
