Menu Close

If-10sin-4-x-15cos-4-x-6-find-the-value-of-27cosec-6-x-8sec-6-x-

Tinku Tara 0 Comments
Pin




Question Number 195253 by Spillover last updated on 28/Jul/23
If 10sin^4 x+15cos^4 x=6. find the value of 27cosec^6 x+8sec^6 x
$${If}\:\mathrm{10sin}\:^{\mathrm{4}} {x}+\mathrm{15cos}\:^{\mathrm{4}} {x}=\mathrm{6}. \\ $$$${find}\:{the}\:{value}\:{of} \\ $$$$\mathrm{27cosec}\:^{\mathrm{6}} {x}+\mathrm{8sec}\:^{\mathrm{6}} {x} \\ $$$$ \\ $$
Commented by Frix last updated on 28/Jul/23
250
$$\mathrm{250} \\ $$
Answered by BaliramKumar last updated on 28/Jul/23
10sin^4 x + 15cos^4 x = 6 ((10)/6)sin^4 x + ((15)/6)cos^4 x = 1 (5/3)sin^4 x + (5/2)cos^4 x = 1 ((5/3)sin^2 x)∙sin^2 x + ((5/2)cos^2 x)∙cos^2 x = 1 ((5/3)sin^2 x) = 1 & ((5/2)cos^2 x) = 1 ((5/3)) = (1/(sin^2 x)) & ((5/2)) = (1/(cos^2 x)) ((5/3)) = cosec^2 x & ((5/2)) = sec^2 x 27cosec^6 x + 8sec^6 x 27((5/3))^3 + 8((5/2))^3 = 27(((125)/(27))) + 8(((125)/8)) 125 + 125 = 250
$$\mathrm{10sin}^{\mathrm{4}} \mathrm{x}\:+\:\mathrm{15cos}^{\mathrm{4}} \mathrm{x}\:=\:\mathrm{6} \\ $$$$\frac{\mathrm{10}}{\mathrm{6}}\mathrm{sin}^{\mathrm{4}} \mathrm{x}\:+\:\frac{\mathrm{15}}{\mathrm{6}}\mathrm{cos}^{\mathrm{4}} \mathrm{x}\:=\:\mathrm{1} \\ $$$$\frac{\mathrm{5}}{\mathrm{3}}\mathrm{sin}^{\mathrm{4}} \mathrm{x}\:+\:\frac{\mathrm{5}}{\mathrm{2}}\mathrm{cos}^{\mathrm{4}} \mathrm{x}\:=\:\mathrm{1} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{3}}\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)\centerdot\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:+\:\left(\frac{\mathrm{5}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)\centerdot\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{1} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{3}}\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)\:=\:\mathrm{1}\:\:\:\&\:\:\left(\frac{\mathrm{5}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)\:=\:\mathrm{1} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{3}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:\:\:\&\:\:\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{3}}\right)\:=\:\mathrm{cosec}^{\mathrm{2}} \mathrm{x}\:\:\:\&\:\:\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\:=\:\mathrm{sec}^{\mathrm{2}} \mathrm{x} \\ $$$$ \\ $$$$\mathrm{27cosec}^{\mathrm{6}} \mathrm{x}\:+\:\mathrm{8sec}^{\mathrm{6}} \mathrm{x} \\ $$$$\mathrm{27}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{3}} \:+\:\mathrm{8}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{3}} \:=\:\mathrm{27}\left(\frac{\mathrm{125}}{\mathrm{27}}\right)\:+\:\mathrm{8}\left(\frac{\mathrm{125}}{\mathrm{8}}\right) \\ $$$$\mathrm{125}\:+\:\mathrm{125}\:=\:\mathrm{250} \\ $$
Commented by Spillover last updated on 03/Aug/23
thank you
$${thank}\:{you} \\ $$
Answered by Spillover last updated on 02/Aug/23
10sin^4 x+15cos^4 x=6 divide by cos^4 x both sides 10tan^4 x+15=6sec^4 x 10tan ^4 x+15=6(1+tan^2 x)^2 4tan^4 x−12tan^2 x+9=0 (2tan^2 x−3)^2 =0 tan^2 x=(3/2) cot^2 x=(2/3) 27cosec^6 x+8sec^6 x 27(1+cot^2 x)^3 +8(1+tan^2 x)^3 but tan^2 x=(3/2) cot^2 x=(2/3) 27(1+(2/3))^3 +8(1+(3/2))^3 27((5/3))^3 +8((5/2))^3 250
$$\mathrm{10sin}\:^{\mathrm{4}} {x}+\mathrm{15cos}\:^{\mathrm{4}} {x}=\mathrm{6} \\ $$$${divide}\:{by}\:{cos}^{\mathrm{4}} {x}\:{both}\:{sides} \\ $$$$\mathrm{10tan}\:^{\mathrm{4}} {x}+\mathrm{15}=\mathrm{6sec}\:^{\mathrm{4}} {x} \\ $$$$\mathrm{10tan}\overset{\mathrm{4}} {\:}{x}+\mathrm{15}=\mathrm{6}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} \\ $$$$\mathrm{4tan}\:^{\mathrm{4}} {x}−\mathrm{12tan}\:^{\mathrm{2}} {x}+\mathrm{9}=\mathrm{0} \\ $$$$\left(\mathrm{2tan}\:^{\mathrm{2}} {x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cot}\:^{\mathrm{2}} {x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{27cosec}\:^{\mathrm{6}} {x}+\mathrm{8sec}\:^{\mathrm{6}} {x} \\ $$$$\mathrm{27}\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} +\mathrm{8}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} \\ $$$${but}\:\:\mathrm{tan}\:^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cot}\:^{\mathrm{2}} {x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{27}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} +\mathrm{8}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\mathrm{27}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{3}} +\mathrm{8}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\mathrm{250} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *