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If 15tan^2+4sec^2=23,then find (sec+cosec)^2-sin^2

If 15tan^2+4sec^2=23,then find (sec+cosec)^2-sin^2
 

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4 Answers

Sunil Raikwar
askIITians Faculty 45 Points
7 years ago
15tan^2 x +4(tan^2 x+1)=23
= 19tan^2 x +4 = 23
= 19tan^2 x = 19
= tan^2 x =1
= tanx= 1
therefore
secx=root2
cosecx=root2
sinx=1/root2

(secx+cosecx)^2-sin^2 x =(2root2)^2-1/2=8-1/2=15/2


Paramjeet Singh Khalsa
20 Points
4 years ago
15 tan²∅ + 4 sec ²∅ = 23 [ tan²∅ = sec²∅ - 1 ]15 [ sec²∅ - 1 ] + 4 sec ²∅ = 2315 sec²∅ - 15 + 4 sec²∅ = 2315 sec²∅ + 4 sec²∅ = 23 + 1519 sec²∅ = 38sec²∅ = 38/19sec²∅ = 2sec∅ = √2 { sec 45° = √2 }sec∅ = sec 45°∅ = 45°( sec 45° + cosec 45° )² - sin² 45°( √2 + √2 )² - ( 1/√2 )²( 2√2 )² - 1/28 - 1/2 = 8 - 0.5 = 7.5
Paramjeet Singh Khalsa
20 Points
4 years ago
15 tan²∅ + 4 sec ²∅ = 23. [ tan²∅ = sec²∅ - 1 ]. 15 [ sec²∅ - 1 ] + 4 sec ²∅ = 23. 15 sec²∅ - 15 + 4 sec²∅ = 23. 15 sec²∅ + 4 sec²∅ = 23 + 15. 19 sec²∅ = 38. sec²∅ = 38/19. sec²∅ = 2. sec∅ = √2. { sec 45° = √2 }. sec∅ = sec 45°. ∅ = 45°. ( sec 45° + cosec 45° )² - sin² 45°. ( √2 + √2 )² - ( 1/√2 )². ( 2√2 )² - 1/2. 8 - 1/2 = 8 - 0.5 = 7.5.
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

15tan^2 + 4sec^2 = 23
15tan^2 x + 4(tan^2 x+1)=23
19tan^2 x + 4 = 23
19tan^2 x = 19
tan^2 x =1
tanx = 1
therefore sec x = root2
cosec x = root2
sinx = 1/root2
(secx+cosecx)^2-sin^2 x
(2root2)^2-1/2
8-1/2
15/2

Thanks and Regards

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