sin⁡(θ+α)cos⁡(θ−α)=1−M1+M then Tan⁡π4−θ⋅Tan⁡π4−α=

sin⁡(θ+α)cos⁡(θ−α)=1−M1+Msinθ+α+cosθ-αsinθ+α-cosθ-α=1-M+1+M1-M-1-M ⇒sinθ+α+sinπ2-θ-αsinθ+α-sinπ2-θ-α=2-2M ⇒2sinθ+α+π2-θ-α2cosθ+α-π2+θ-α22cosθ+α+π2-θ-α2sinθ+α-π2+θ-α2=-1M ⇒tanπ4+αcotθ-π4=-1M ⇒M=-1tanπ4+α-tanπ4-θ ∴M=tanπ4-αtanπ4-θ

sin⁡(θ+α)cos⁡(θ−α)=1−M1+M then Tan⁡π4−θ⋅Tan⁡π4−α=

sin⁡(θ+α)cos⁡(θ−α)=1−M1+Msinθ+α+cosθ-αsinθ+α-cosθ-α=1-M+1+M1-M-1-M ⇒sinθ+α+sinπ2-θ-αsinθ+α-sinπ2-θ-α=2-2M ⇒2sinθ+α+π2-θ-α2cosθ+α-π2+θ-α22cosθ+α+π2-θ-α2sinθ+α-π2+θ-α2=-1M ⇒tanπ4+αcotθ-π4=-1M ⇒M=-1tanπ4+α-tanπ4-θ ∴M=tanπ4-αtanπ4-θ

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 M$

$\frac{1}{M}$

$\frac{- 1}{M}$

$M$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{\sin (θ + α)}{\cos (θ - α)} = \frac{1 - M}{1 + M}$

$\frac{\sin (θ + α) + \cos (θ - α)}{\sin (θ + α) - \cos (θ - α)} = \frac{1 - M + 1 + M}{1 - M - 1 - M} \Rightarrow \frac{\sin (θ + α) + \sin (\frac{π}{2} - (θ - α))}{\sin (θ + α) - \sin (\frac{π}{2} - (θ - α))} = \frac{2}{- 2 M} \Rightarrow \frac{2 \sin (\frac{θ + α + \frac{π}{2} - (θ - α)}{2}) \cos (\frac{θ + α - \frac{π}{2} + (θ - α)}{2})}{2 \cos (\frac{θ + α + \frac{π}{2} - (θ - α)}{2}) \sin (\frac{θ + α - \frac{π}{2} + (θ - α)}{2})} = \frac{- 1}{M} \Rightarrow \tan (\frac{π}{4} + α) c o t (θ - \frac{π}{4}) = \frac{- 1}{M} \Rightarrow M = - \frac{1}{\tan (\frac{π}{4} + α)} (- \tan (\frac{π}{4} - θ)) ∴ M = \tan (\frac{π}{4} - α) \tan (\frac{π}{4} - θ)$