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Question Number 91689 by john santu last updated on 02/May/20

sec^2 1^o +sec^2 2^o +sec^2 3^o +...+sec^2 89^o

$$\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}^{{o}} +\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}^{{o}} +\mathrm{sec}\:^{\mathrm{2}} \mathrm{3}^{{o}} +...+\mathrm{sec}\:^{\mathrm{2}} \mathrm{89}^{{o}} \\ $$

Commented by Prithwish Sen 1 last updated on 02/May/20

sec^2 a+cose^2 a= ((sin^2 a+cos^2 a)/(sin^2 a cos^2 a)) = cosec^2 a sec^2 a

$$\mathrm{sec}^{\mathrm{2}} \mathrm{a}+\mathrm{cose}^{\mathrm{2}} \mathrm{a}=\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{a}+\mathrm{cos}^{\mathrm{2}} \mathrm{a}}{\mathrm{sin}^{\mathrm{2}} \mathrm{a}\:\mathrm{cos}^{\mathrm{2}} \mathrm{a}}\:=\:\mathrm{cosec}^{\mathrm{2}} \mathrm{a}\:\mathrm{sec}^{\mathrm{2}} \mathrm{a} \\ $$

Commented by Prithwish Sen 1 last updated on 02/May/20

sec^2 1cosec^2 1+.........+sec^2 44cosec^2 44+2  =4(cosec^2 2+cosec^2 4+...+cosec^2 88)+2  =16(cosec^2 4+cosec^2 8+...+cosec^2 60+...+cosrc^2 88)+2  =16(cosec^2 2+cosec^2 4+...+cosec^2 42+cosec^2 44)+2  continue..

$$\mathrm{sec}^{\mathrm{2}} \mathrm{1cosec}^{\mathrm{2}} \mathrm{1}+.........+\mathrm{sec}^{\mathrm{2}} \mathrm{44cosec}^{\mathrm{2}} \mathrm{44}+\mathrm{2} \\ $$$$=\mathrm{4}\left(\mathrm{cosec}^{\mathrm{2}} \mathrm{2}+\mathrm{cosec}^{\mathrm{2}} \mathrm{4}+...+\mathrm{cosec}^{\mathrm{2}} \mathrm{88}\right)+\mathrm{2} \\ $$$$=\mathrm{16}\left(\mathrm{cosec}^{\mathrm{2}} \mathrm{4}+\mathrm{cosec}^{\mathrm{2}} \mathrm{8}+...+\mathrm{cosec}^{\mathrm{2}} \mathrm{60}+...+\mathrm{cosrc}^{\mathrm{2}} \mathrm{88}\right)+\mathrm{2} \\ $$$$=\mathrm{16}\left(\mathrm{cosec}^{\mathrm{2}} \mathrm{2}+\mathrm{cosec}^{\mathrm{2}} \mathrm{4}+...+\mathrm{cosec}^{\mathrm{2}} \mathrm{42}+\mathrm{cosec}^{\mathrm{2}} \mathrm{44}\right)+\mathrm{2} \\ $$$$\mathrm{continue}.. \\ $$

Commented by jagoll last updated on 02/May/20

why sec^2 1 cosec^2 1 ?  sec^2 89 = cosec^2 1 ⇒ sec^2 1+sec^2 89  = sec^2 1+cosec^2 1 sir

$${why}\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}\:\mathrm{cosec}\:^{\mathrm{2}} \mathrm{1}\:? \\ $$$$\mathrm{sec}\:^{\mathrm{2}} \mathrm{89}\:=\:\mathrm{cosec}\:^{\mathrm{2}} \mathrm{1}\:\Rightarrow\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}+\mathrm{sec}\:^{\mathrm{2}} \mathrm{89} \\ $$$$=\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}+\mathrm{cosec}\:^{\mathrm{2}} \mathrm{1}\:{sir} \\ $$

Commented by jagoll last updated on 02/May/20

oouh yes. thank you sir

$${oouh}\:{yes}.\:{thank}\:{you}\:{sir} \\ $$

Commented by jagoll last updated on 02/May/20

but how to get 4sin^2  2 ?  sec^2 1 cosec^2 1 = (4/((2sin 1 cos 1)^2 )) =  (4/(sin^2  2)) = 4 cosec^2  2 sir?

$${but}\:{how}\:{to}\:{get}\:\mathrm{4sin}\:^{\mathrm{2}} \:\mathrm{2}\:? \\ $$$$\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}\:\mathrm{cosec}\:^{\mathrm{2}} \mathrm{1}\:=\:\frac{\mathrm{4}}{\left(\mathrm{2sin}\:\mathrm{1}\:\mathrm{cos}\:\mathrm{1}\right)^{\mathrm{2}} }\:= \\ $$$$\frac{\mathrm{4}}{\mathrm{sin}\:^{\mathrm{2}} \:\mathrm{2}}\:=\:\mathrm{4}\:\mathrm{cosec}\:^{\mathrm{2}} \:\mathrm{2}\:{sir}? \\ $$

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