Question

pls answer i will for sure follow who answers​

Answer 1

To ProvE :

$\tt \dagger \: \: \: \: \:\dfrac {Sin \, x - tan \, x }{tan \, x .Sin \, x}= \dfrac{Cos\,x - 1 }{Sin\, x}$

QuestioN :

$\tt \dagger \: \: \: \: \: Prove \: the \: Identity :$

$\tt \dagger \: \: \: \: \:\dfrac {Sin \, x - tan \, x }{tan \, x .Sin \, x}= \dfrac{Cos\,x - 1 }{Sin\, x}$

SolutioN :

$\tt \dagger \: \: \: \: \:\dfrac {Sin \, x - tan \, x }{tan \, x .Sin \, x}= \dfrac{Cos\,x - 1 }{Sin\, x}$

☛ Taking LHS.

$\tt : \implies\dfrac {Sin \, x - tan \, x }{tan \, x .Sin \, x}$

✎ We can also write as,

$\tt : \implies\dfrac {Sin \, x }{tan \, x .Sin \, x} - \dfrac{tan \, x }{tan \, x .Sin \, x}$

$\tt : \implies\dfrac { \cancel{Sin \, x }}{tan \, x . \cancel{Sin \, x}} - \dfrac{ \cancel{tan \, x} }{ \cancel{tan \, x} .Sin \, x}$

$\tt : \implies\dfrac {1}{tan \, x} - \dfrac{1 }{Sin \, x}$

$\tt : \implies\dfrac {1}{tan \, x} - \dfrac{1 }{Sin \, x}$

• › Note : 1 / tan x = cot x.

$\tt : \implies Cot\, x - \dfrac{1 }{Sin \, x}$

• › Cot x → Cos x / Sin x.

$\tt : \implies\dfrac {Cos \, x }{Sin \, x} - \dfrac{1 }{Sin \, x}$

$\tt : \implies \dfrac{Cos \, x - 1}{Sin \, x}$

♢ Hence Proved.