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1.

यदि `tan{cos^(-1)(4/5) + tan^(-1)(2/3)}` का आंकिक मान `a/b` हो , तब b का मान होगाA. 6B. 5C. 8D. 10

Answer» Correct Answer - A
माना `cos^(-1)(4/5)= alpha`
` rArr cos alpha= 4/5`
` :. tan alpha = 3/4`
अब `tan{cos^(-1)(4/5)+tan^(-1)(2/3)}`
` = tan(alpha + tan^(-1). 2/3)`
` = (tan alpha+ 2/3)/(1 - tan alpha* 2/3) = 17/6 = a/b` , [दिया है ]
अतः a = 17 तथा b = 6
Discussion
2.

यदि `tan{cos^(-1)(4/5) + tan^(-1)(2/3)}` का आंकिक मान `a/b` हो , तब a का मान होगाA. 16B. 17C. 20D. 15

Answer» Correct Answer - B
माना `cos^(-1)(4/5)= alpha`
` rArr cos alpha= 4/5`
` :. tan alpha = 3/4`
अब `tan{cos^(-1)(4/5)+tan^(-1)(2/3)}`
` = tan(alpha + tan^(-1). 2/3)`
` = (tan alpha+ 2/3)/(1 - tan alpha* 2/3) = 17/6 = a/b` , [दिया है ]
अतः a = 17 तथा b = 6
Discussion
3.

(x, y) के ऐसे कौन- से मान है जो युगपत समीकरणों , ` sin^(-1) x + sin^(-1) y = (2 pi)/3` और `cos^(-1) x - cos^(-1) y = pi/3` को संतुष्ट करते है ?A. `(0,1)`B. `(1//2,1)`C. `(1,1//2)`D. `(sqrt3//2,1)`

Answer» Correct Answer - B
दी गई समीकरणें निम्न है
` sin^(-1)x + sin^(-1) y = (2 pi)/ 3 ` ...(i)
तथा ` cos^(-1) x - cos^(-1) y = pi/3 `...(ii)
` rArr pi/2 - sin^(-1) x - pi/2 + sin ^(-1) y = pi/3`
` rArr - sin^(-1) x + sin^(-1) y = pi/3 ` ....(iii)
समी (i) व(iii) को हल करने पर ,
` 2 sin^(-1) y = pi ` तथा ` 2 sin^(-1) x = pi//3`
` rArr sin^(-1) y = pi//2 ` तथा ` sin^(-1) x = pi//6`
` rArr x = 1//2`
तथा ` y = 1 `
` rArr (x,y) = (1//2, 1)`
Discussion
4.

`sin^(-1)((1-x^(2))/(1+x^(2)))`का मान है ?A. `2 tan ^(-1) x`B. ` pi + tan^(-1) x`C. ` pi/2 - 2 tan ^(-1) x`D. इनमे से कोई नहीं

Answer» Correct Answer - C
माना `x = tan theta`
` :. sin^(-1)((1-x^(2))/(1+x^(2))) = sin^(-1)((1-tan^(2)theta)/(1+tan^(2) theta))`
` = sin^(-1) (cos 2theta) = sin^(-1)[sin(pi/2 - 2 theta)]`
` = pi/2 - 2 theta = pi/2 - 2 tan^(-1) x`
Discussion
5.

यदि `ax + b {sec(tan^(-1)x)} = c` तथा `ay + b {sec(tan^(-1)y)} = c`, तब x + y का मान होगाA. `(2ac)/(a^(2)-b^(2))`B. `(c^(2)-b^(2))/(a^(2)-b^(2))`C. ` (c^(2)-b^(2))/(a^(2)+b^(2))`D. इनमें से कोई नहीं

Answer» Correct Answer - A
मान लीजिए `tan^(-1) x = alpha` तथा `tan^(-1) y = beta`
तब , `a tan alpha + b sec alpha = c ` ...(i)
तथा `a tan beta+ b sec beta = c ` ...(ii)
समी (i) और (ii) से स्पष्ट है कि समीकरण `a tan theta + b sec theta = c ` के मूल `alpha`तथा `beta`है ।
अब ,`a tan theta + b sec theta = c `
` rArr b sec theta = c - a tan theta`
` rArr b^(2) sec^(2) theta = c^(2) - 2ac tan theta + a^(2) tan^(2) theta`
` rArr b^(2) + b^(2) tan^(2) theta = c^(2) - 2ac tan theta + a^(2) tan ^(2) theta`
` rArr (a^(2) - b^(2)) tan^(2) theta - 2ac tan theta + c^(2) - b^(2) = 0`
` rArr tan alpha+ tan beta = x + y = (2ac)/(a^(2) - b^(2))`
तथा `tan alpha* tan beta = xy = (c^(2) - b^(2))/(a^(2) - b^(2))`
` (x+y)/(1-xy)=((2ac)/(a^(2) -b^(2)))/(1-(c^(2)-b^(2))/(a^(2) -b^(2)))=(2ac)/(a^(2) -c^(2))` तथा `(1 + xy)/(x+y) = (1+(c^(2)-b^(2))/(a^(2)-b^(2)))/((2ac)/(a^(2)-b^(2)))`
` = (a^(2) + c^(2) - 2b^(2))/(2ac)`
Discussion
6.

यदि `ax + b {sec(tan^(-1)x)} = c` तथा `ay + b {sec(tan^(-1)y)} = c`, तब xy का मान होगाA. `(2ab)/(a^(2)-b^(2))`B. `(c^(2)-b^(2))/(a^(2) -b^(2))`C. `(c^(2)-b^(2))/(a^(2)+b^(2))`D. इनमें से कोई नहीं

Answer» Correct Answer - B
मान लीजिए `tan^(-1) x = alpha` तथा `tan^(-1) y = beta`
तब , `a tan alpha + b sec alpha = c ` ...(i)
तथा `a tan beta+ b sec beta = c ` ...(ii)
समी (i) और (ii) से स्पष्ट है कि समीकरण `a tan theta + b sec theta = c ` के मूल `alpha`तथा `beta`है ।
अब ,`a tan theta + b sec theta = c `
` rArr b sec theta = c - a tan theta`
` rArr b^(2) sec^(2) theta = c^(2) - 2ac tan theta + a^(2) tan^(2) theta`
` rArr b^(2) + b^(2) tan^(2) theta = c^(2) - 2ac tan theta + a^(2) tan ^(2) theta`
` rArr (a^(2) - b^(2)) tan^(2) theta - 2ac tan theta + c^(2) - b^(2) = 0`
` rArr tan alpha+ tan beta = x + y = (2ac)/(a^(2) - b^(2))`
तथा `tan alpha* tan beta = xy = (c^(2) - b^(2))/(a^(2) - b^(2))`
` (x+y)/(1-xy)=((2ac)/(a^(2) -b^(2)))/(1-(c^(2)-b^(2))/(a^(2) -b^(2)))=(2ac)/(a^(2) -c^(2))` तथा `(1 + xy)/(x+y) = (1+(c^(2)-b^(2))/(a^(2)-b^(2)))/((2ac)/(a^(2)-b^(2)))`
` = (a^(2) + c^(2) - 2b^(2))/(2ac)`
Discussion
7.

यदि `ax + b {sec(tan^(-1)x)} = c` तथा `ay + b {sec(tan^(-1)y)} = c`, तब ` (x+y)/(1-xy)` का मान होगाA. `(2ab)/(1^(2)-c^(2))`B. `(2ac)/(a^(2)-c^(2))`C. `(c^(2)-b^(2))/(a^(2)+b^(2))`D. इनमें से कोई नहीं

Answer» Correct Answer - B
मान लीजिए `tan^(-1) x = alpha` तथा `tan^(-1) y = beta`
तब , `a tan alpha + b sec alpha = c ` ...(i)
तथा `a tan beta+ b sec beta = c ` ...(ii)
समी (i) और (ii) से स्पष्ट है कि समीकरण `a tan theta + b sec theta = c ` के मूल `alpha`तथा `beta`है ।
अब ,`a tan theta + b sec theta = c `
` rArr b sec theta = c - a tan theta`
` rArr b^(2) sec^(2) theta = c^(2) - 2ac tan theta + a^(2) tan^(2) theta`
` rArr b^(2) + b^(2) tan^(2) theta = c^(2) - 2ac tan theta + a^(2) tan ^(2) theta`
` rArr (a^(2) - b^(2)) tan^(2) theta - 2ac tan theta + c^(2) - b^(2) = 0`
` rArr tan alpha+ tan beta = x + y = (2ac)/(a^(2) - b^(2))`
तथा `tan alpha* tan beta = xy = (c^(2) - b^(2))/(a^(2) - b^(2))`
` (x+y)/(1-xy)=((2ac)/(a^(2) -b^(2)))/(1-(c^(2)-b^(2))/(a^(2) -b^(2)))=(2ac)/(a^(2) -c^(2))` तथा `(1 + xy)/(x+y) = (1+(c^(2)-b^(2))/(a^(2)-b^(2)))/((2ac)/(a^(2)-b^(2)))`
` = (a^(2) + c^(2) - 2b^(2))/(2ac)`
Discussion
8.

` tan^(-1)(1/2) + tan^(-1)(1/3) ` किसके बराबर है ?A. `pi/2`B. `pi/3`C. `pi/4`D. `pi/6`

Answer» Correct Answer - C
`tan^(-1). 1/2 + tan^(-1). 1/3 = tan^(-1) ((1/2+1/3)/(1-1/6))`
` = tan^(-1) (tan. pi/4) = pi/4`
Discussion
9.

` sin^(-1) 3/5 + sin ^(-1) 4/5` किसके बराबर है ?A. `pi/2`B. `pi/3`C. `pi/4`D. `pi/6`

Answer» Correct Answer - A
दिया है , ` sin^(-1)(3/5) + sin^(-1)(4/5)`
माना ` sin^(-1)(4/5) = theta rArr sin theta = (4/5)`
` rArr cos theta = sqrt(1-sin^(2) theta)`
` rArr cos theta = 3/5`
तथा ` sin ^(-1)(4/5) = theta = cos^(-1) (3/5)`
` :. sin^(-1)(3/5)+ sin^(-1)(4/5)`
` = sin^(-1) (3/5) + cos^(-1) (3/5) = pi/2`
Discussion
10.

मान लीजिए` x = 4 tan^(-1)(1/5) , y = tan^(-1) (1/70)` और ` z = tan^(-1)(1/99)` है । x - y किसके बराबर है ?A. `tan^(-1)((828)/845)`B. ` tan^(-1)((8287)/(8450))`C. ` tan^(-1)((8281)/(8450))`D. ` tan^(-1)((8287)/(8471))`

Answer» Correct Answer - C
` x - y = tan^(-1)(120/119) - tan^(-1)(1/70)`
` = tan^(-1)((120/119-1/70)/(1+120/119xx1/70))`
` = tan^(-1)((8281)/(8450))`
Discussion
11.

मान लीजिए` x = 4 tan^(-1)(1/5) , y = tan^(-1) (1/70)` और ` z = tan^(-1)(1/99)` है । x - y + z किसके बराबर है ?A. ` pi/2`B. `pi/3`C. ` pi/6`D. ` pi/4`

Answer» Correct Answer - D
` x - y + z = tan^(-1)((8281)/(8450))+tan^(-1)(1/99)`
` = tan^(-1) ((828269)/(828269)) = pi/4`
Discussion
12.

यदि x और y धनात्मक है और `xy gt 1` है , तो `tan^(-1) x + tan^(-1) y` किसके तुल्य है ?A. ` tan^(-1)((x+y)/(1-xy))`B. ` pi + tan^(-1)((x+y)/(1-xy))`C. `pi - tan^(-1)((x+y)/(1-xy))`D. ` tan^(-1)((x-y)/(1+xy))`

Answer» Correct Answer - B
हम जानते है कि , यदि x तथा y धनात्मक है
अर्थात ` x gt 0, y gt 0` तथा ` xy gt 1`, तब
` tan^(-1) x + tan^(-1) y = pi + tan^(-1)((x+y)/(1-xy))`
Discussion
13.

`sec^(2) {tan^(-1)(5/11)}` का क्या मान है ?A. 121/96B. 217/921C. 146/121D. 267/121

Answer» Correct Answer - C
Discussion
14.

मान लीजिए` x = 4 tan^(-1)(1/5) , y = tan^(-1) (1/70)` और ` z = tan^(-1)(1/99)` है । x किसके बराबर है ?A. ` tan^(-1)(60/119)`B. `tan^(-1)(120/119)`C. ` tan^(-1)(90/169)`D. ` tan^(-1)(170/169)`

Answer» Correct Answer - B
`x = 4 tan^(-1) (1/5) = 2*2 tan^(-1) (1/5)`
` = 2 tan^(1)((2/5)/(1-1/25))`
` = 2 tan^(-1)(5/12)`
` = tan^(-1)((10/12)/(1-25/144))`
` = tan^(-1) (120/119)`
Discussion
15.

यदि ` 4 sin^(-1) x + cos^(-1) x = pi` है , तो x का मान होगाA. 1B. `1/2`C. `-1/2`D. `1/4`

Answer» Correct Answer - B
दिया है ,`4 sin^(-1) x + cos^(-1) x = pi`
` rArr 3 sin^(-1) x + sin^(-1) x + cos^(-1) x = pi`
हम जानते है कि,
` sin^(-1) x + cos^(-1) x = pi/2`
` :. 3 sin^(-1) x = pi - pi/2 = pi/2`
` rArr sin^(-1) x = pi/6 rArr x = sin . pi/6`
अतः ` x = 1//2`
Discussion