\(\frac{2\,tan\,3^\circ}{1 + tan^2\,30^\circ} = \)
(2 tan 3°)/(1 + tan2 30°) =
(A) sin 30°
(B) cos 60°
(C) tan 60°
(D) sin 60°
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Prove that `tan alpha+2 tan 2 alpha+2^(2)tan^(2)alpha+..............+2^(n-1)tan2^(n-1)alpha+2^(@)cot2^(@)alpha=cotalpha`
We know `tan theta=cot theta-2 cot 2 theta`. Putting `theta=alpha,2alpha,2^(2)alpha,....................` in (i), we get `tan alpha=(cot alpha-2 cot 2alpha)` `2(tan2alpha)=2(cot 2 alpha-2 cot 2 ^(2)alpha)` `2^(2)(tan2^(2)alpha)=2^(2)(cot2^(2)alpha-2cot2^(3)alpha)` `2^(n-1)(tan2^(n-1)alpha)=2^(n-1)(cot2^(n-1)alpha-2cos2^(n)alpha)` Adding, `tan alpha+2tan...
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If `[(1,-tan theta),(tan theta,1)][(1,tan theta),(-tan theta,1)]^(-1)=[(a,-b),(b,a)],` then
We have, `[(1,tan theta),(-tan theta,1)]^(-1) =1/(1+ tan^(2) theta) [(1,- tan theta),(tan theta,1)]` `implies [(a,-b),(b,a)]=cos^(2) theta [(1,-tan theta),(tan theta,1)][(1,-tan theta),(tan theta,1)]` `=cos^(2) theta [(1-tan^(2) theta,-2 tan theta),(2 tan theta,1-tan^(2) theta)]` `=[(cos...
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If `0<a_1<a_2<ddot<a_n ,` then prove that `tan^(-1)((a_1x-y)/(x+a_1y))+tan^(-1)((a_2-a_1)/(1+a_2a+1))+tan^(-1)((a_3-a_2)/(1+a_3a_2))++tan^(-1)((a_n-a_
Here, `tan^(_1) ((a_(1) x- y)/(a + a_(1) y)) = tan^(-1) ((a_(1) - (y)/(x))/(1 + a_(1) (y)/(x))) = tan^(-1) a_(1) - tan^(-1) (y)/(x)` `tan^(-1) ((a_(2) -a_(1))/(1 + a_(2) a_(1))) = tan^(-1)...
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Find the principal values of the following (i) `tan^(-1) (tan. (2pi)/(3))` (ii) `tan^(-1) (tan (-6))`
(i) `tan^(-1) (tan.(2pi)/(3)) != (2pi)/(3), " as " (2pi)/(3)` does not lie between `- (pi)/(2) and (pi)/(2)` Now, `tan^(-1) (tan.(2pi)/(3)) = tan^(-1) (tan(pi - (pi)/(3)))` `= tan^(-1) (- tan. (pi)/(3))`...
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Prove that : `2 tan^(-1) (cosec tan^(-1) x - tan cot^(-1) x) = tan^(-1) x`
Correct Answer - C `2 tan^(-1 (cosec tan^(-1) x - tan cot^(-1) x)` `= 2 tan^(-1) [cosec {cosec^(-1) (sqrt(1 + x^(2)))/(x)} - tan^(-1) {tan^(-1) ((1)/(x))}]` `= 2 tan^(-1) [sqrt((1 + x^(2))/(x))...
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If `u=cot^-1 sqrt(tanalpha)-tan^-1 sqrt(tan alpha),` then `tan(pi/4-u/2)` is equal to (a) `sqrt(tan alpha)` (b) `sqrt(cos alpha)` (c) `tan alpha` (d)
Correct Answer - A Let `sqrt(tan alpha) = tan x`. Then `u = cot^(-1) (tan x) - tan^(-1) (tan x)` `= (pi)/(2) - x -x = (pi)/(2) - 2x` or `2x...
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`theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) " then " tan theta=`
Correct Answer - A `theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) = tan^(-1).(2 tan^(2) theta - (1)/(3) tan theta)/(1 + (1)/(3) tan^(3) theta)` `rArr tan theta =...
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The value of `tan^(-1).(4)/(7) + tan^(-1).(4)/(19) + tan^(-1).(4)/(39) + tan^(-1).(4)/(67) ... oo` equals
Correct Answer - B Let us find the value of general term of the series `S = 7 + 19 + 39 + 67 + .. + T_(n)` `(S = "...
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निम्नलिखित समीकरणों को हल कीजिए - `tan^(-1)(x-1)+tan^(-1)x+tan^(-1)(x+1)=tan^(-1)3x.`
Correct Answer - `0,+-(1)/(2)`
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15. Which of the following is correct? (1) \( \tan 1>\tan 2 \) (2) \( \tan 2>\tan 1 \) (3) \( \sin 1\cos 2 \)
Answer: (1) tan1>tan2
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