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Let prod(r=1)^51 tan(pi/3(1+(3^r)/(3^(5...

Let `prod_(r=1)^51 tan(pi/3(1+(3^r)/(3^(50)-1))=prod_(r=1)^51 cot(pi/3(1+(3^r)/(3^(50)-1))])` On solving equation we get, `1-3 tan^2 (1+(3^r)/(3^(50)-1))=a/(bk-1),(a,b in I)` then value of `(a-b)` is equal

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prod_(r=1)^( Let )tan((pi)/(3)(1+(3^(r))/(3^(50)-1))=prod_(r=1)^(51)cot((pi)/(3)(1+(3^(r))/(3^(50)-1))]) On solving equation we get,1-3tan^(2)(1+(3^(r))/(3^(50)-1))=(a)/(bk-1),(a,b in I) then value of (a-b) is equal

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ARIHANT MATHS-TRIGONOMETRIC FUNCTIONS AND IDENTITIES-Exercise (Questions Asked In Previous 13 Years Exam)
  1. Let prod(r=1)^51 tan(pi/3(1+(3^r)/(3^(50)-1))=prod(r=1)^51 cot(pi/3(1...

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  2. If alpha and beta are non-zero real number such that 2(cos beta-cos al...

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  3. Let -1/6 < theta < -pi/12 Suppose alpha1 and beta1, are the root...

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  4. The value of sum(k=1)^(13) (1)/(sin(pi/4 + ((k-1)pi)/(6))sin(pi/4 + (k...

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  5. Let f : (-1, 1) -> R be such that f(cos4theta) = 2/(2-sec^2theta for t...

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  6. The number of all possible values of theta, where 0 lt theta lt pi, f...

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  7. For 0 lt theta lt pi/2, the solution (s) of sum(m=1)^(6) cosec (the...

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  8. If sin^ 4 x/2+cos^4 x/3 =1/5 then

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  9. Let theta in (0,pi/4) and t1=(tan theta)^(tan theta), t2=(tan theta)6(...

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  10. cos(alpha-beta)=1a n dcos(alpha+beta)=l/e , where alpha,betamu in [-pi...

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  11. If 5(tan^2x - cos^2x)=2cos 2x + 9, then the value of cos4x is

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  12. Let fk(x) = 1/k(sin^k x + cos^k x) where x in RR and k gt= 1. Then f4(...

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  13. The expression (tanA)/(1-cotA)+(cotA)/(1-tanA) can be written as (1) s...

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  14. If a Delta PQR " if" 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P =1 , ...

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  15. If A = sin^2x + cos^4 x, then for all real x :

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  16. Let cos (alpha+beta) = 4/5 and sin(alpha-beta)=5/13 where 0<= alpha,...

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  17. If cosalpha+cosbeta+cosgamma=0=sinalpha+sinbeta+singamma, then which...

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  18. A triangular park is enclosed on two sides by a fence and on the third...

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  19. If 0 lt x lt pi and cos x + sin x = 1/2, then tan x is

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  20. In Delta PQR , /R=pi/4, tan(P/3), tan(Q/3) are the roots of the equati...

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