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The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal to
- a)(2n+1)π/4
- b)(2n+1)π/6
- c)(2n+1)π/2
- d)(2n+1)π/3
Correct answer is option 'B'. Can you explain this answer?
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The general solution of the equation tan2θ.tanθ=1 for n&is...
The given equation is tan2x = 1.
To find the general solution, we can first find the values of x that satisfy the equation tan2x = 1.
Using the identity tan2x = (2tanx)/(1-tan^2x), we can rewrite the equation as:
(2tanx)/(1-tan^2x) = 1
Multiplying both sides by (1-tan^2x), we get:
2tanx = (1-tan^2x)
Rearranging the equation, we have:
tan^2x + 2tanx - 1 = 0
Now, we can solve this quadratic equation for tanx.
Using the quadratic formula, we have:
tanx = (-2 ± √(2^2 - 4(1)(-1))) / 2(1)
tanx = (-2 ± √(4 + 4)) / 2
tanx = (-2 ± √8) / 2
tanx = (-2 ± 2√2) / 2
tanx = -1 ± √2
So, the values of tanx that satisfy the equation are -1 + √2 and -1 - √2.
To find the general solution, we can use the fact that tan has a period of π.
So, the general solution is:
x = (n + 1/4)π, where n is an integer.
Therefore, the general solution of the equation tan2x = 1 is x = (n + 1/4)π, where n is an integer.
To find the general solution, we can first find the values of x that satisfy the equation tan2x = 1.
Using the identity tan2x = (2tanx)/(1-tan^2x), we can rewrite the equation as:
(2tanx)/(1-tan^2x) = 1
Multiplying both sides by (1-tan^2x), we get:
2tanx = (1-tan^2x)
Rearranging the equation, we have:
tan^2x + 2tanx - 1 = 0
Now, we can solve this quadratic equation for tanx.
Using the quadratic formula, we have:
tanx = (-2 ± √(2^2 - 4(1)(-1))) / 2(1)
tanx = (-2 ± √(4 + 4)) / 2
tanx = (-2 ± √8) / 2
tanx = (-2 ± 2√2) / 2
tanx = -1 ± √2
So, the values of tanx that satisfy the equation are -1 + √2 and -1 - √2.
To find the general solution, we can use the fact that tan has a period of π.
So, the general solution is:
x = (n + 1/4)π, where n is an integer.
Therefore, the general solution of the equation tan2x = 1 is x = (n + 1/4)π, where n is an integer.
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The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal toa)(2n+1)π/4b)(2n+1)π/6c)(2n+1)π/2d)(2n+1)π/3Correct answer is option 'B'. Can you explain this answer?
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The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal toa)(2n+1)π/4b)(2n+1)π/6c)(2n+1)π/2d)(2n+1)π/3Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal toa)(2n+1)π/4b)(2n+1)π/6c)(2n+1)π/2d)(2n+1)π/3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal toa)(2n+1)π/4b)(2n+1)π/6c)(2n+1)π/2d)(2n+1)π/3Correct answer is option 'B'. Can you explain this answer?.
The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal toa)(2n+1)π/4b)(2n+1)π/6c)(2n+1)π/2d)(2n+1)π/3Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal toa)(2n+1)π/4b)(2n+1)π/6c)(2n+1)π/2d)(2n+1)π/3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The general solution of the equation tan2θ.tanθ=1 for n∈I is, θ is equal toa)(2n+1)π/4b)(2n+1)π/6c)(2n+1)π/2d)(2n+1)π/3Correct answer is option 'B'. Can you explain this answer?.
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