If 2 tan- - - - tan- prove that - - 2 tan - - -

We need to prove β 2tan(α β) Thus β 2 [tanα tanβ1+tanα.tanβ]=0 Taking LHS: 11 tanα.tanβ [β+tanα 2tanα+2tanβ] 11 tanα.tanβ [β tanα+2tanβ] ...

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Trigonometric Ratios of Compound Angles

If 2 tanβ +...

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If $2 tan β + cot β = tan α$ , prove that $cot β = 2 tan (α - β) .$

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2 tan β + cot β = tan α

cot β = 2 tan (α - β)

If 2 $t a n α = 3 t a n β$ , prove that $t a n (α - β) = \frac{s i n 2 β}{5 - c o s 2 β}$

tan α + b tan β = (a + b) tan (\frac{α + β}{2})

α \neq β

a cos β = b cos α

tan (α + β) + tan (α - β) = \frac{sin 2 α}{1 - {sin}^{2} α - {sin}^{2} β}

tan β = \frac{n sin α cos α}{1 - n {sin}^{2} α}

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