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Question-215696

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Question Number 215696 by BaliramKumar last updated on 15/Jan/25
Answered by MATHEMATICSAM last updated on 15/Jan/25
sec^4 x − cosec^4 x − 2sec^2 x + 2cosec^2 x = ((63)/8) ⇒ sec^2 x(sec^2 x − 2) + cosec^2 x(2 − cosec^2 x) = ((63)/8) ⇒ (1 + tan^2 x)(tan^2 x − 1) + (1 + cot^2 x)(1 − cot^2 x) = ((63)/8) ⇒ (tan^4 x − 1) + (1 − cot^4 x) = ((63)/8) ⇒ tan^4 x − 1 + 1 − cot^4 x = ((63)/8) ⇒ tan^4 x − cot^4 x = ((63)/8)
$$\mathrm{sec}^{\mathrm{4}} \mathrm{x}\:−\:\mathrm{cosec}^{\mathrm{4}} \mathrm{x}\:−\:\mathrm{2sec}^{\mathrm{2}} \mathrm{x}\:+\:\mathrm{2cosec}^{\mathrm{2}} \mathrm{x}\:=\:\frac{\mathrm{63}}{\mathrm{8}} \\ $$$$\Rightarrow\:\mathrm{sec}^{\mathrm{2}} {x}\left(\mathrm{sec}^{\mathrm{2}} {x}\:−\:\mathrm{2}\right)\:+\:\mathrm{cosec}^{\mathrm{2}} {x}\left(\mathrm{2}\:−\:\mathrm{cosec}^{\mathrm{2}} \mathrm{x}\right)\:=\:\frac{\mathrm{63}}{\mathrm{8}} \\ $$$$\Rightarrow\:\left(\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{2}} {x}\right)\left(\mathrm{tan}^{\mathrm{2}} {x}\:−\:\mathrm{1}\right)\:+\:\left(\mathrm{1}\:+\:\mathrm{cot}^{\mathrm{2}} {x}\right)\left(\mathrm{1}\:−\:\mathrm{cot}^{\mathrm{2}} {x}\right)\:=\:\frac{\mathrm{63}}{\mathrm{8}} \\ $$$$\Rightarrow\:\left(\mathrm{tan}^{\mathrm{4}} {x}\:−\:\mathrm{1}\right)\:+\:\left(\mathrm{1}\:−\:\mathrm{cot}^{\mathrm{4}} {x}\right)\:=\:\frac{\mathrm{63}}{\mathrm{8}} \\ $$$$\Rightarrow\:\mathrm{tan}^{\mathrm{4}} {x}\:−\:\mathrm{1}\:+\:\mathrm{1}\:−\:\mathrm{cot}^{\mathrm{4}} {x}\:=\:\frac{\mathrm{63}}{\mathrm{8}} \\ $$$$\Rightarrow\:\mathrm{tan}^{\mathrm{4}} {x}\:−\:\mathrm{cot}^{\mathrm{4}} {x}\:=\:\frac{\mathrm{63}}{\mathrm{8}} \\ $$
Commented by MATHEMATICSAM last updated on 15/Jan/25
I got something like this
$$\mathrm{I}\:\mathrm{got}\:\mathrm{something}\:\mathrm{like}\:\mathrm{this} \\ $$
Answered by A5T last updated on 15/Jan/25
sec^4 x−cosec^4 x−2sec^2 x+2cosec^2 x=((63)/8) ⇒(sec^2 x−1)^2 −(cosec^2 x−1)^2 =((63)/8) sec^2 x=tan^2 x+1; cosec^2 x=1+cot^2 x ⇒tan^4 x−(1/(tan^4 x))=((63)/8)⇒8tan^8 x−63tan^4 x−8=0 tan^4 x=((63+_− (√(63^2 −4(−8)8)))/(16))>0⇒tan^4 x=((63+65)/(16))=8 ⇒tan^2 x=2(√2)⇒cot^2 x=(1/(2(√2)))=((√2)/4) ⇒tan^2 x+cot^2 x=((9(√2))/4)=(9/(2(√2)))
$$\mathrm{sec}^{\mathrm{4}} \mathrm{x}−\mathrm{cosec}^{\mathrm{4}} \mathrm{x}−\mathrm{2sec}^{\mathrm{2}} \mathrm{x}+\mathrm{2cosec}^{\mathrm{2}} \mathrm{x}=\frac{\mathrm{63}}{\mathrm{8}} \\ $$$$\Rightarrow\left(\mathrm{sec}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{cosec}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{63}}{\mathrm{8}} \\ $$$$\mathrm{sec}^{\mathrm{2}} \mathrm{x}=\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{1};\:\mathrm{cosec}^{\mathrm{2}} \mathrm{x}=\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{4}} \mathrm{x}−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{4}} \mathrm{x}}=\frac{\mathrm{63}}{\mathrm{8}}\Rightarrow\mathrm{8tan}^{\mathrm{8}} \mathrm{x}−\mathrm{63tan}^{\mathrm{4}} \mathrm{x}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{tan}^{\mathrm{4}} \mathrm{x}=\frac{\mathrm{63}\underset{−} {+}\sqrt{\mathrm{63}^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{8}\right)\mathrm{8}}}{\mathrm{16}}>\mathrm{0}\Rightarrow\mathrm{tan}^{\mathrm{4}} \mathrm{x}=\frac{\mathrm{63}+\mathrm{65}}{\mathrm{16}}=\mathrm{8} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{2}} \mathrm{x}=\mathrm{2}\sqrt{\mathrm{2}}\Rightarrow\mathrm{cot}^{\mathrm{2}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{cot}^{\mathrm{2}} \mathrm{x}=\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Commented by BaliramKumar last updated on 15/Jan/25
also tan^4 x−cot^4 x = ((63)/8) tan^4 x − (1/(tan^4 x)) = 8 − (1/8) tan^4 x=8 tan^2 x=(√8) ( (√8) + (1/( (√8))) = (9/(2(√2))))
$${also}\:\:\:{tan}^{\mathrm{4}} {x}−{cot}^{\mathrm{4}} {x}\:=\:\frac{\mathrm{63}}{\mathrm{8}} \\ $$$${tan}^{\mathrm{4}} {x}\:−\:\frac{\mathrm{1}}{{tan}^{\mathrm{4}} {x}}\:=\:\mathrm{8}\:−\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${tan}^{\mathrm{4}} {x}=\mathrm{8} \\ $$$${tan}^{\mathrm{2}} {x}=\sqrt{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:\:\sqrt{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}}}\:=\:\frac{\mathrm{9}}{\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$$ \\ $$

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