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Question Number 126610 by bramlexs22 last updated on 22/Dec/20

 Given  { ((cos x+sin x=p)),((sec x+csc x = q  )) :}   p≠q . Find ((cos^2 x)/(2−2sin^2 x−csc^2  x )) .

$$\:{Given}\:\begin{cases}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}={p}}\\{\mathrm{sec}\:{x}+\mathrm{csc}\:{x}\:=\:{q}\:\:}\end{cases} \\ $$$$\:{p}\neq{q}\:.\:{Find}\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{2}−\mathrm{2sin}\:^{\mathrm{2}} {x}−\mathrm{cs}{c}^{\mathrm{2}} \:{x}\:}\:. \\ $$

Answered by liberty last updated on 22/Dec/20

 q = (((cos x+sin x)/(cos x sin x)))⇒q^2  = (p^2 /(cos^2 x sin^2 x))  cos^2 x sin^2 x = (p^2 /(q^2  )) ...(i)  ⇒ ((cos^2 x)/(2cos^2 x−(1/(sin^2 x)))) = ((cos^2 x sin^2 x)/(2cos^2 x sin^2 x−1))                                     =  (( ((p^2 /q^2 )))/((((2p^2 )/q^2 )−1))) = (p^2 /(2p^2 −q^2 ))

$$\:{q}\:=\:\left(\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}\right)\Rightarrow{q}^{\mathrm{2}} \:=\:\frac{{p}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{sin}\:^{\mathrm{2}} {x}} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{sin}\:^{\mathrm{2}} {x}\:=\:\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} \:}\:...\left({i}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{2cos}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} {x}}}\:=\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{2cos}\:^{\mathrm{2}} {x}\:\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\:\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\right)}{\left(\frac{\mathrm{2}{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }−\mathrm{1}\right)}\:=\:\frac{{p}^{\mathrm{2}} }{\mathrm{2}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} } \\ $$

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