If tan 2 -tan-1 and tan-tan y-1 then

The correct options are A (1 + tanα)(1 + tan(β2)) = 2sin nbsp;(α + γ) C tan nbsp;(α + β) = tanγ D tan 2γtanβ = 1Given, tan 2α×tanβ = 1 ⇒ 2α = π2 β⇒ 2α + ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Equations

If tan 2 α·ta...

Question

If $tan 2 α \cdot tan β = 1$ and $tan α \cdot tan γ = 1$ then

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

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tan α + tan β = tan γ

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tan (α + β) = tan γ

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tan 2 γ tan β = - 1

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Solution

The correct options are
A $(1 + tan α) (1 + tan (\frac{β}{2})) = 2 sin (α + γ)$
C $tan (α + β) = tan γ$
D $tan 2 γ tan β = - 1$
Given,
$tan 2 α \times tan β = 1 \Rightarrow 2 α = \frac{π}{2} - β \Rightarrow 2 α + β = \frac{π}{2} . . . (1)$

$tan α \times tan γ = 1 \Rightarrow α = \frac{π}{2} - γ \Rightarrow α + γ = \frac{π}{2} . . . (2)$

$(1 + tan α) (1 + tan (\frac{π}{4} - α))$
$= (1 + tan α) \times \frac{2}{(1 + tan α)} = 2 = R H S$

From $(1)$
$α + β = \frac{π}{2} - α$
$∴ tan (α + β) = cot α = cot (\frac{π}{2} - γ) = tan γ$

$α + β = \frac{π}{2} \Rightarrow 2 γ = π - 2 α \Rightarrow tan 2 γ = - tan 2 α$
and $β = \frac{π}{2} - 2 α \Rightarrow tan β = cot 2 α$
$∴ tan 2 γ \cdot tan β = - 1 = R H S$

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If tan2α⋅tanβ=1 and tanα⋅tanγ=1 then

If $tan 2 α \cdot tan β = 1$ and $tan α \cdot tan γ = 1$ then