Identities involving trig functions are listed below Pythagorean Identities sin 2 θ cos 2 θ = 1 tan 2 θ 1 = sec 2 θ cot 2 θ 1 = csc 2 θ Reciprocal IdentitiesSec^2 xtan^2 x=3 tan x2tan^2 x 1=3 tan x2 tan^2 x 2 tan^2 x3 tan x1=0 By solving it we are getting two values of tan x that are 1/2 and 1 But the value of 1 will be rejected because tan^1 (1)=45° and for 45° both sin x = tan x and it is provided in the question that the value of sin x is not equal to cos x hence the value of tan x =1/2=0512/2/ Transcript Ex 33, 22 Prove that cot 𝑥 cot 2𝑥 – cot 2𝑥 cot 3𝑥 – cot 3𝑥 cot 𝑥 = 1 Taking LHS cot x cot 2x – cot 2x cot 3x – cot 3x cot x = cot x cot 2x – cot 3x (cot 2x cot x) = cot x cot 2x – cot (2x x) (cot 2x cot x) = cot x cot 2x – ( (cot 2x cot x − 1)/ (cot x cot 2x)) (cot 2x cot x) = cot x cot 2x – (cot 2x cot x – 1) =
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Tan 2x is equal to minus cot x + 5 by 3-Tanx = 3cotxtanx = 3×1/tanx(cotx = t/tanx)tan2x = 3tanx = 3 So the value of x will be 60o because the value of tan 60 is 3Get NCERT Exemplar Solutions for Class 11 Chapter Trigonometric Functions here BeTrainedin has solved each questions of NCERT Exemplar very thoroughly to help the students in solving any question from the book with a team of well experianced subject matter experts Practice Trigonometric Functions questions and become a master of concepts
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Tan(x) = square root of 3 Take the inverse tangent of both sides of the equation to extract from inside the tangent The exact value of is The tangent function is positive in the first and third quadrants To find the second solution,13/2/ Example 22 Solve tan 2x = – cot (x" " 𝜋/3) tan 2x = –cot (𝑥" " 𝜋/3) We need to make both in terms of tan Rough tan (90° θ) = –cot θ –cot θ = tan (90° θ) –cot θ = tan (𝜋/2 " θ" ) Replacing θ by x 𝜋/3 –cot ("x " 𝜋/3) = tan (𝜋/2 " x " 𝜋/3) tan 2Plus Gama equals 4/3 And this is a compound angle identity for 10 So using the identity from our formula booklet, this is the same as 10 um, data plus 10 gamma over one minus 10th ada 10 gamma equal 4/3 On we can fill in the sights of 10 data was X 10 gamma is ah one minus X and then one minus 10 Thade is X and tan gamma was one minus X
19/9/ Evaluate ∫(2 tan x – 3 cot x)² dx (a) 4tan x – cot x – 25x C (b) 4 tan x – 9 cot x – 25x C (c) – 4 tan x 9 cot x 25x C (d) 4 tan x 9 cot x 25x C Answer Answer (b) 4 tan x – 9 cot x – 25x CINVERSE TRIGONOMETRIC FUNCTIONS 23 Therefore, tan(cos–1x) = 1–cos θ 21– tanθ = cosθ x x = Hence 2 –1 8 1– 8 17 15 tan cos = 17 8 8 17 = Example 11 Find the value of –1 –59/6/21 For all real values of x, cot x – 2 cot 2x is equal to A tan 2x B tan x C cot 3x D None of these
The integral ∫ π 4 3 π 4 d x 1 cos x is equal to Integrals 8 The area (in sq units) of the region { ( x, y) x ≥ 0, x y ≤ 3, x 2 ≤ 4 y and y ≤ 1 x } is Application of Integrals 9 If ( 2 sin x) d y d x ( y 1) cos x = 0 and y ( 0) = 1, then y ( π 2) is equal to Differential Equations 1017/2/18 Explanation tan2x −cotx = 0 tan2x = cotx tan2x ⋅ tanx = 1 sin2x ⋅ sinx cos2x ⋅ cosx = 1 1 2 ⋅ (cosx − cos3x) 1 2 ⋅ (cos3 cosx) = 1 cosx − cos3x cos3x cosx = 1 cosx − cos3x = cos3x cosx 2cos3x = 0Now, sin 2x = 2tan x/ (1 tan^2 x) , putting tan x = /1 sin 2x = 2 (/1)/ {1 (/1)^2} or, sin 2x = 2 (/1)/2 or, sin 2x = /1 Answer Basically tan theta = cot theta is only possible when theta = 45 degree or (pie/4) radians or odd multiple of that So there are 2 cases
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Introduction to Tan double angle formula let's look at trigonometric formulae also called as the double angle formulae having double angles Derive Double Angle Formulae for Tan 2 Theta \(Tan 2x =\frac{2tan x}{1tan^{2}x} \) let's recall the addition formulaAnd time was of Minour Square root of three Is actually equal to minus off by by three As for God inverse of court and worst of minus square root of three, it is april two by minus Bye Bye six, which is equal to 5, 5 x six So these are the Values of the expressions that we knowSolve x divided by nine is less than or equal to negative three for x Show your work math When x is divided by 3,the remainder is zIn terms of z, which of the following could be equal to x ?
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12/5/21 Ex 22, 21 Find value of tan−1 √3 – cot−1 (−√3) Finding tan−1 √𝟑 Let y = tan−1 √3 tan y = √3 tan y = tan (𝝅/𝟑) ∴ y = 𝝅/𝟑 Since range of tan1 is ( (−π)/2,π/2) Hence, Principal Value is 𝝅/𝟑 (Since tan 𝜋/3 = √3) ∴ tan−1 √3 = π/3 Finding cot−1 (−√𝟑) Let x = cot−1 (√3) x = 𝜋 − cot−1 (√3) x = 𝜋 − 𝛑/𝟔 x30/3/16 How do I solve this?A z3 b 3z c 3z d 6z e 92z please answer and explain math Find the exact value of 1 tan 75 degrees 2 Cos 15 degrees 3
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If y=\sin ^{1} 2 x\tan ^{1} 3 x\cos ^{1} 2 x\cot ^{1} 3 x then (a) y^{\prime}(0)=0 (b) y_{5}=y_{6} (c) y_{2}=y_{A} (d) y_{y}(0)=1 Video Transcript {'transcript' 'in this single correct answer type question given differential equation is that is why dash plus sign off X plus Y overdue is equal to sign off X minus light overdueSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreCot Inverse Calculator Are you looking for a smart tool that calculates the cotangent inverse of a real number or fraction within no time?Then, you have arrived the correct place and our calculator is the best tool that you are looking for The main aim of our Cot Inverse Calculator is to calculate the cotangent inverse of numbers simply and quickly
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Given,y = tan21 cos−1( 35 )Let, x= cos−1( 35 )⇒ cosx = 35 ∴ y = tan 21 xy = tan 2x y = 1cosx1−cosx = 1 35 1− 35 = 3 5 3− 5 rationalizing the factor, we get,y = 23− 54) cot2(2x) Answer (3) 2 cot 2x Solution cot x – tan x = (1/tan x) – tan x = (1 – tan 2 x)/tan x Multiplying and dividing by 2, = 2 (1 – tan 2 x)/2 tan x Using the formula tan 2A = 2 tan A/ (1 – tan 2 A), = 2/tan 2x Explanation tan2x −cot2x = sin2(x) cos2(x) − cos2(x) sin2(x) = sin4(x) − cos4(x) cos2(x)sin2(x) = (sin2x cos2x)(sin2x − cos2x) (cos(x)sin(x))2 = (1)( − cos(2x)) 1 4sin2(2x) = − 4 cos(2x) sin2(2x) = − 4csc(2x)cot(2x) Final Answer
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General Solution of Tan 5 X = Cot 2 X isTan x cot y/tan x cot y= tan y cot x Math Factorthen use fundamental identities to simplify the expression below and determine which of the following is not equivalent cot^2 a * tan^2 a cot^2 a A csc^ 2 alpha B1/ sin^ 2 alpha C1/ 1cos^ 2 alpha Dsec^ 2 alphaTan 2 θ = sec 2 θ − 1 The square of tan function equals to the subtraction of one from the square of secant function is called the tan squared formula It is also called as the square of tan function identity
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The answer is =2csc(2x) We need tanx=sinx/cosx cotx=cosx/sinx cos^2xsin^2s=1 sin2x=2sinxcosx cscx=1/sinx Therefore, cotxtanx=cosx/sinxsinx/cosx =(cos^2xsin^2x)/(sinxcosx) =1/(sinxcosx) =2/sin(2x) =2csc(2x)If cot2x =cot(x−y)cot(x −z), then cot2x is equal to (where x =±π/4 )12/5/ X equal to tan a minus tan B and Y equal to COT B minus cot a then 1/x1/y equal to
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Tan^2x= 2tanxsinx My work so far tan^2x 2tanxsinx=0 tanx(tanx 2sinx)=0 Then the solutions are TanX=0 and sinX/cosX = 2 sin X Divide through by sinX we have to check this later to see if allowed (ie sinX math Prove the identity 1cos2x/sin2x = 1tanx/1cotx I simplified the RS to tanxClick here👆to get an answer to your question ️ If sin^4x 2 cos^4x 3 = 1 5 , then Join / Login maths If 2 s i n 4 x 3 c o s 4 x = 5 1 , then;BITSAT 16 If tan ( cot x) = cot ( tan x), then sin 2x is equal to (A) (2/(2n1)π) (B) (4/(2n1)π) (2/n(n1)π) (D) (4/n(n1)π) Ch BITSAT 16 If tan ( cot x) = cot ( tan x), then sin 2x is equal to (A) (2/(2n1)π) (B) (4/(2n1)π) (2/n(n1)π) (D) (4/n(n1)π) Ch Tardigrade
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25/7/19 The value of is option 2) Stepbystep explanation We have, To find, the value of ∴ Using identity, ∵ Using trigonometric identity, Hence, the value of is option 2)Solve the equation 4 tan^2 x 12 sec x 1 =0, for 0 degrees is less than or equal to x is less than or equal to 360 degrees trig tan^2theta6=sec^2theta5 calculus a bridge is 10 m above a canal A motorboat going 3 m/sec passes under the center of the bridge at the same instant that a woman walking 2 m/sec reaches that point30/5/ If cos^2x cos^4x =1 , then the value of tan^4x cot^4x tan^2x cot^2x is equal to ?
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Trigonometry Solve for x cot (x10)=tan (4x) cot (x − 10) = tan (4x) cot ( x 10) = tan ( 4 x) Move all terms containing x x to the left side of the equation Tap for more steps Subtract tan ( 4 x) tan ( 4 x) from both sides of the equation cot ( x − 10) − tan ( 4 x) = 0 cot ( x 10) tan ( 4 x) = 0 Simplify the left side of theWe have, 3tan−1xcot−1x= π⇒ 2tan−1 x(tan−1xcot−1x)= π⇒ 2tan−1 x 2π = π, ∵ tan−1xcot−1x = 2π is an identity⇒ 2tan−1 x= π− 2π = 2π ⇒ tan−1x = 4π ∴ x= tan 4π =123/3/18 If 3 tan1 x cot1 x = π, then x equals (a) 0 (b) 1 (c) 1 (d) 1/2 Questions from AIIMS 10 1 An alternating current i in an inductance coil varies with time according to
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Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutor28/8/ If cot x equal to minus 3 by 5 x in 2 quadrant then find sinx by 2 cos x by 2 tan x by2 1 Log in Join now 1 Log in Join now Ask your question janvi3295 janvi3295 Math Secondary School #tan 2x = (2tan x)/(1 tan^2 x)# (Trig identity) #2/(tan 2x) = (1 tan^2 x)/tan x = 1/(tan x) tan^2 x/(tan x) =# #= cot x tan x#
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1) when the expression 4x^23x8 is divided by xa, the remainder is 2 find the value of a 2) the polynomial 3x^3mx^2nx5 leaves a remainder of 128 when divided by x3 and a remainder of 4 when divided by x1 calculate the remainder when the Algebra 2Math\sin^2x\cos^2x=1/math math\implies\dfrac{\sin^2x}{\cos^2x}\dfrac{\cos^2x}{\cos^2x}=\dfrac{1}{\cos^2x}/math math\implies\left(\dfrac{\sin x}{\cos xGiven \tan x \cot x = 3 and x is in first quadrant Find \sin x Given tanxcotx = 3 and x is in first quadrant Find sinx Method 1 Recall that \sin (2x) = 2\sin (x)\cos (x) From what you have we have \sin (x) \cos (x) = \dfrac13 \implies \sin (2x) = \dfrac23 Since x lies in the first quadrant, 2x
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For the first term in equation (1) on the right, we'll use the basic trigonometric identity cot x = 1/tan x as follows (2) = {sin² x/ 1 (1/tan x)} cos² x/ (1 tan x) (3) = {sin² x/ (tan x 1)/tan x)} cos² x/ (1 tan x) Dividing fractions in the first term, we getA tan 2 x = 3 29/1/18 This is readily derived directly from the definition of the basic trigonometric functions sin and cos and Pythagoras's Theorem Divide both side by cos2x and we get sin2x cos2x cos2x cos2x ≡ 1 cos2x ∴ tan2x 1 ≡ sec2x ∴ tan2x ≡ sec2x − 1 Confirming that the result is an identity
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