If sin-1sin-2-cos-1cos-2-1-0- where -1-2-0-2 - then the value of 1-tan-1-41-tan-2-4 is

Sinθ _1 sinθ_2 cosθ_1cosθ_2=1 ⇒cos(θ_1+θ_2)= 1 ⇒ nbsp;θ_1+θ_2=π as θ_1+θ_2∈ (0,2π) nbsp; ⇒θ_1+θ_24=π4 ⇒ nbsp;θ_24=π4 θ_14 ∴ nbsp; (1+tanθ_14)(1+tanθ_2 ...

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Byju's Answer

Standard XII

Mathematics

Cos(A+B)Cos(A-B)

If sinθ1sinθ2...

Question

If $sin θ_{1} sin θ_{2} - cos θ_{1} cos θ_{2} - 1 = 0,$ where $θ_{1} + θ_{2} \in (0, 2 π),$ then the value of $(1 + tan \frac{θ_{1}}{4}) (1 + tan \frac{θ_{2}}{4})$ is

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Solution

$sin θ_{1} sin θ_{2} - cos θ_{1} cos θ_{2} = 1$
$\Rightarrow cos (θ_{1} + θ_{2}) = - 1$
$\Rightarrow θ_{1} + θ_{2} = π$ as $θ_{1} + θ_{2} \in (0, 2 π)$
$\Rightarrow \frac{θ_{1} + θ_{2}}{4} = \frac{π}{4}$
$\Rightarrow \frac{θ_{2}}{4} = \frac{π}{4} - \frac{θ_{1}}{4}$

$∴ (1 + tan \frac{θ_{1}}{4}) (1 + tan \frac{θ_{2}}{4})$
$= (1 + tan \frac{θ_{1}}{4}) [1 + tan (\frac{π}{4} - \frac{θ_{1}}{4})]$
$= (1 + tan \frac{θ_{1}}{4}) ⎡ ⎢ ⎢ ⎢ ⎣ \frac{2}{1 + tan \frac{θ_{1}}{4}} ⎤ ⎥ ⎥ ⎥ ⎦ = 2$

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