If sinθ+αcosθ-α=1-M1+M then Tanπ4-θ.Tanπ4-α=

sinθ+αsin90+α-θ=1-M1+Msinθ+α+sin90+α-θsinθ+α-sin90+α-θ=1-M+1+M1-M-1-M2sin45+αcos45-θ-2cos45+αsin45-θ=2-2M tan45+αcot45-θ=1Mcot45-αcot45-θ=1Mtan45-α.tan45-θ=M

If sinθ+αcosθ-α=1-M1+M then Tanπ4-θ.Tanπ4-α=

sinθ+αsin90+α-θ=1-M1+Msinθ+α+sin90+α-θsinθ+α-sin90+α-θ=1-M+1+M1-M-1-M2sin45+αcos45-θ-2cos45+αsin45-θ=2-2M tan45+αcot45-θ=1Mcot45-αcot45-θ=1Mtan45-α.tan45-θ=M

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If $\frac{\sin (θ + α)}{\cos (θ - α)} = \frac{1 - M}{1 + M} then Tan (\frac{π}{4} - θ) . Tan (\frac{π}{4} - α) =$

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$\frac{1}{M}$

$- \frac{1}{M}$

answer is B.

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Detailed Solution

$\frac{\sin (θ + α)}{\sin (90 + α - θ)} = \frac{1 - M}{1 + M}$
$\frac{\sin (θ + α) + \sin (90 + α - θ)}{\sin (θ + α) - \sin (90 + α - θ)} = \frac{1 - M + 1 + M}{1 - M - 1 - M}$
$\frac{2 \sin (45 + α) \cos (45 - θ)}{- 2 \cos (45 + α) \sin (45 - θ)} = \frac{2}{- 2 M} \frac{}{}$
$\tan (45 + α) \cot (45 - θ) = \frac{1}{M}$
$\cot (45 - α) \cot (45 - θ) = \frac{1}{M}$
$\tan (45 - α) . \tan (45 - θ) = M$