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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
निम्न समीकरणों को हल कीजिए - `3 tan^(-1)""(1)/(2+sqrt(3))-tan^(-1)""(1)/(x)=tan^(-1)""(1)/(3)` |
| Answer» Correct Answer - 2 | |
| 102. |
निम्न समीकरणों को हल कीजिएः `"tan"^(-1)(sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))=alpha` |
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Answer» Correct Answer - `+-sqrtsin2theta` |
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| 103. |
निम्न समीकरणों को हल कीजिए - `cos(tan^(-1)x)=sin(cot^(-1).(3)/(4)).` |
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Answer» `cos(tan^(-1)x)=sin(cot^(-1).(3)/(4))` `rArr sin((pi)/(2)-tan^(-1)x)=sin(cot^(-1).(3)/(4))` `" "[because cos theta=sin((pi)/(2)-theta)]` दोनों पक्षों की तुलना करने पर, `(pi)/(2)-tan^(-1)x=cot^(-1).(3)/(4)` `rArr" "tan^(-1)x+cot^(-1).(3)/(4)=(pi)/(2)` हम जानते हैं कि `tan^(-1)x+cot^(-1)x=(pi)/(2x in R` `therefore" "cot^(-1).(3)/(4)=cot^(-1)x` `rArr" "x=(3)/(4).` |
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| 104. |
मूल्यांकन कीजिए - `tan^(-1)(sqrt3)-sec^(-1)(-2)+"cosec"^(-1)((2)/(sqrt3))`. |
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Answer» `tan^(-1)(sqrt3)=(pi)/(3)" "`[उदाहरण 1 (iii ) से] `sec^(-1)(-2)=(2pi)/(3)" "` [उदाहरण 2 (iv ) से ] माना `"cosec"^(-1)((2)/(sqrt3))=theta` `rArr" cosec"theta =(2)/(sqrt3)." "theta in [-(pi)/(2),(pi)/(2)]-{0}` `rArr" cosec"theta="cosec"(pi)/(3)` `rArr" "theta=(pi)/(3) "cosec" [-(pi)/(2),(pi)/(2)]-{0}` `therefore tan^(-1)sqrt3-sec^(-1)(-2)+"cosec"^(-1)((2)/(sqrt3))` `=(pi)/(3)-(2pi)/(3)+(pi)/(3)` = 0. |
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| 105. |
`cos^(-1)(cos""(7pi)/(6))` का मान बराबर हैA. `(7pi)/(6)`B. `(5pi)/(6)`C. `(pi)/(3)`D. `(pi)/(6)` |
| Answer» Correct Answer - B | |
| 106. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए : `sec^(-1)((2)/(sqrt3))` |
| Answer» `sec^(-1)((2)/(sqrt3))=sec^(-1)(sec""(pi)/(6))=(pi)/(6)` | |
| 107. |
यदि `sin^(-1)(1-x)-2 sin^(-1)x=(pi)/(2) ` तो x का मान बराबर है :A. `0,(1)/(2)`B. `1,(1)/(2)`C. `0`D. `(1)/(2)` |
| Answer» Correct Answer - C | |
| 108. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए - `sec^(-1)(-(2)/(3)sqrt3)` |
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Answer» माना `sec^(-1)(-(2)/(3)sqrt3)=theta` `rArr" "sec^(-1)(-(2)/(sqrt3))=theta, theta in [0,pi]-{(pi)/(2)}` `rArr" "sec theta=-(2)/(sqrt3)` `rArr" "sec theta=sec(pi-(pi)/(6))` `rArr" "sec theta=sec.(5pi)/(6)` `rArr" "theta=(5pi)/(6)` अतः `sec^(-1)(-(2)/(3)sqrt3)` का मुख्य मान `(5pi)/(6)` है। |
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| 109. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए - `sec^(-1)((2)/(sqrt3))` |
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Answer» हम जानते हैं कि `sec^(-1)` की मुख्य शाखा का परिसर `[0,pi]-{(pi)/(2)}` होता है। माना `" "sec^(-1)((2)/(sqrt3))=theta` `rArr" "sec theta=(2)/(sqrt3), theta in [0,pi]-{(pi)/(2)}` `rArr" "sec theta=sec.(pi)/(6)` `rArr" "theta=(pi)/(6)` अतः `sec^(-1)((2)/(sqrt3))` का मुख्य मान `(pi)/(6)` है। |
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| 110. |
`3sin^(-1)x=sin^(-1)(3x-4x^(3)),x"in[-(1)/(2),(1)/(2)]` |
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Answer» माना `sin^(-1)x=theta` `impliesx=sintheta` `R.H.S=sin^(-1)(3x-4x^(3))` `=sin^(-1)(3sintheta-4sin^(3)theta)` `=sin^(-1)(sin3theta)` `=3theta=3sin^(-1)x` =L.H.S |
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| 111. |
`sin[sin^(-1)(-(1)/(2))+(pi)/(3)]` का मान है-A. `-sqrt(3)/(2)`B. `-(1)/(2)`C. `(1)/(2)`D. `sqrt(3)/(2)` |
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Answer» Correct Answer - C |
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| 112. |
` " cosec" (sin^(-1)x+ cos^(-1) x)` का मुख्य मान ज्ञात कीजिए । |
| Answer» Correct Answer - 1 | |
| 113. |
`cosec^(-1) (2)` का मुख्य मान निकालें |
| Answer» Correct Answer - `(pi)/(6)` | |
| 114. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए : `cosec^(-1)(-sqrt2)` |
| Answer» Correct Answer - `(-pi)/(4)` | |
| 115. |
यदि `tan^(-1)""((x-2)/(x-4))+tan^(-1)""((x+2)/(x+4))=(pi)/(4)` तब x का मान ज्ञात कीजिए । |
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Answer» दी गयी समीकरण को हम इस प्रकार लिख सकते है । `tan^(-1)""((x-2)/(x-4))=(pi)/(4)-tan^(-1)""((x+2)/(x+4))` `=tan^(-1)1-tan^(-1)""((x+2)/(x+4))=tan^(-1)""[(1-(x+2)/(x+4))/(1+(x+2)/(x+4))]` `=tan^(-1)""((2)/(2x+6))=tan^(-1)""((1)/(x+3))` इसलिए , `(x-2)/(x-4)=(1)/(x+3)` `implies (x-2)(x+3)=x-4` `implies x^(2)+x-6=x-4` `implies x^(2)=2 implies x=pm sqrt(2)` |
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| 116. |
यदि `sin[cot^(-1)(1+x)]=cos(tan^(-1)x)` तब x=A. `(1)/(2)`B. `-(1)/(2) `C. `0`D. 1 |
| Answer» Correct Answer - B | |
| 117. |
`cos^(-1)[cos""(5pi)/(3)]+sin^(-1)[sin""(5pi)/(3)]` का मान है -A. `(pi)/(2)`B. `(5pi)/(3)`C. `(10pi)/(3)`D. `0` |
| Answer» Correct Answer - D | |
| 118. |
निम्न समीकरणों को हल कीजिए - ` sin^(-1)""(5)/(x)+sin^(-1)""(12)/(x)=(pi)/(2)` |
| Answer» Correct Answer - 13 | |
| 119. |
निम्नलिखित समीकरणों को हल कीजिए - `tan^(-1)((x-1)/(x+2))+tan^(-1)((x+1)/(x+2))=(pi)/(4).` |
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Answer» दिया गया समीकरण है- `tan^(-1)((x-1)/(x-2))+tan^(-1)((x+1)/(x+2))=(pi)/(4)` `rArr" "tan^(-1)(((x-1)/(x-2)+(x+1)/(x+2))/(1-((x-1)/(x-2))((x+1)/(x+2))))=(pi)/(4),` `[because tan^(-1)x+tan^(-1)y=tan^(-1)((x+y)/(1-xy))]` `rArr tan^(-1)[(((x-1)(x+2)+(x+1)(x-2))/((x-2)(x+2)))/(((x-2)(x+2)-(x-1)(x+1))/((x-2)(x+2)))]=(pi)/(4)` `rArr tan^(-1)[(x^(2)+2x-x-2+x^(2)+2x+x-2)/((x^(2)-4)-(x^(2)-1))]=(pi)/(4)` `rArr" "tan^(-1)((2x^(2)-4)/(-3))=(pi)/(4)` `rArr" "(2x^(2)-4)/(-3)=tan.(pi)/(4)` `rArr" "(2x^(2)-4)/(-3)=1` ltbr `rArr" "2x^(2)-4=-3` `rArr" "2x^(2)=1` `rArr" "x=pm(1)/(sqrt2).` |
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| 120. |
निम्नलिखित का मान ज्ञात कीजिए : `tan^(-1)sqrt(3)-sec^(-1)(-2) `का मान बराबर हैA. `pi`B. `-(pi)/(3)`C. `(pi)/(3)`D. `(2pi)/(3)` |
| Answer» Correct Answer - B | |
| 121. |
`tan^(-1)sqrt3-sec^(-1)(-2)` का मान बराबर है :A. `pi`B. `-(pi)/(3)`C. `(pi)/(3)`D. `(2pi)/(3)` |
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Answer» Correct Answer - B `tan^(-1)sqrt3-sec^(-1)(-2)` `=tan^-1(tan""(pi)/(3))-[pi-sec^(-1)2]` `=(pi)/(3)-pi+sec^(-1)(sec""(pi)/(3))` `(pi)/(3)-pi+(pi)/(3)=-(pi)/(3)` |
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| 122. |
`sin(2sin^(-1)""(4)/(5))` का मान है-A. `(40)/(41)`B. `(9)/(41)`C. `(16)/(25)`D. इनमे से कोई नहीं। |
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Answer» Correct Answer - A |
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| 123. |
यदि `0 lt x lt 1` हो तब ` sqrt(1+x^(2))[{x cos(cot^(-1)x)+sin(cot^(-1)x)}^(2)-1]^((1)/(2))`=...A. xB. `x^(2)`C. `x sqrt(1+x^(2))`D. `sqrt(1+x^(2))` |
| Answer» Correct Answer - C | |
| 124. |
यदि ` tan^(-1)x+tan^(-1)y=(pi)/(4)` तब ` x+y+xy` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - 1 | |
| 125. |
`sin(sec^(-1)x + " cosec"^(-1) x)` का मुख्य मान ज्ञात कीजिए । |
| Answer» Correct Answer - 1 | |
| 126. |
निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए : ` tan""(1)/(2)[sin^(-1)""(2x)/(1+x^(2))+cos^(-1)""(1-y^(2))/(1+y^(2))],|x| lt 1 , y gt 0` तथा `xy lt 1` |
| Answer» Correct Answer - `(x+y)/(1-xy)` | |
| 127. |
यदि `tan^(-1)((3)/(4))+tan^(-1)((1)/(k))=(pi)/(4)` हो, तो k का मान ज्ञात कीजिए । |
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Answer» दिया है- `tan^(-1)((3)/(4))+tan^(-1)((1)/(k))=(pi)/(4)` `rArr tan^(-1)((3)/(4))+tan^(-1)((1)/(k))=tan^(-1)1` `rArr" "tan^(-1)((1)/(k))=tan^(-1)1-tan^(-1)((3)/(4))` `=tan^(-1)((1)/(k))=tan^(-1)((1-(3)/(4))/(1+(3)/(4)))=tan^(-1)((1-(3)/(4))/(1+(3)/(4)))=tan^(-1)(((4-3)/(4))/((4+3)/(4)))` `rArr tan^(-1)((1)/(k))=tan^(-1)((1)/(7))` `rArr" "(1)/(k)=(1)/(7)` `rArr" "k=7.` |
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| 128. |
निम्नलिखित समीकरणों को हल कीजिए - `tan^(-1)2x+tan^(-1)3x=(pi)/(4).` |
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Answer» दिया गया समीकरण है - `tan^(-1)2x+tan^(-1)3x=(pi)/(4)` `rArr" "tan^(-1)((2x+3x)/(1-(2x)/(3x)))=(pi)/(4)` `rArr" "tan^(-1)((5x)/(1-6x^(2)))=(pi)/(4)` `rArr" "(5x)/(1-6x^(2))=tan.(pi)/(4)` `rArr" "(5x)/(1-6x^(2))=1` `rArr" "6x^(2)+5x-1=0` `rArr" "6x^(2)+6x-x-1=0` `rArr 6x(x+1)-1(x+1)=0` `rArr" "(6x-1)(x+1)=0` `rArr" "x=(1)/(6), x=-1` ltMbrgt चूँकि `x=-1` दिए गये समीकरण को संतुष्ट नहीं करता है। `therefore x=(1)/(6).` |
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| 129. |
निम्नलिखित समीकरणों को हल कीजिए - `tan^(-1)x+2cot^(-1)x=(2pi)/(3).` |
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Answer» दिया गया समीकरण है - `tan^(-1)x+2cot^(-1)x=(2pi)/(3)` `rArr" "tan^(-1)x+2tan^(-1)((1)/(x))=(2pi)/(3),` `" "[because cot^(-1)x=tan^(-1).(1)/(x)]` `rArr" "tan^(-1)x+tan^(-1)((2xx+(1)/(x))/(1-(1)/(x^(2))))=(2pi)/(3)` `rArr" "tan^(-1)x+tan^(-1)(((2)/(x))/((x^(2)-1)/(x^(2))))=(2pi)/(3)` `rArr" "tan^(-1)x+tan^(-1)((2x)/(x^(2)-1))=(2pi)/(3)` `rArr" "tan^(-1)((x+(2x)/(x^(2)-1))/(1-x xx(2x)/(x^(2)-1)))=(2pi)/(3)` `rArr" "tan^(-1)((x^(3)+x)/(-1-x^(2)))=(2pi)/(3)` `rArr" "(x^(3)+x)/(-1-x^(2))=tan((2pi)/(3))` `rArr" "-(x^(3)+x)/(1+x^(2))tan(pi-(pi)/(3))` `rArr" "-(x^(3)-x)/(1+x^(2))=-tan.(pi)/(3)` `rArr" "(x(1+x^(2)))/(1+x^(2))=sqrt3.` `rArr" "x=sqrt3.` |
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| 130. |
`sin(tan^(-1)x), |x| lt 1` बराबर होता है :A. `(x)/(sqrt(1-x^(2)))`B. `(1)/(sqrt(1-x^(2)))`C. `(1)/(sqrt(1+x^(2)))`D. `(x)/(sqrt(1+x^(2)))` |
| Answer» Correct Answer - D | |
| 131. |
`tan^(-1)""(1-x)/(1+x)=(1)/(2)tan^(-1)x,(xgt0)` |
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Answer» `tan^(-1)""(1-x)/(1+x)=(1)/(2)tan^(-1)x,(xgt0)` `impliestan^(-1)-tan^(-1)x=(1)/(2)tan^(-1)x` `implies(pi)/(4)=(1)/(2)tan^(-1)x+tan^(-1)x` `implies(3)/(2)tan^(-1)x=(pi)/(4)` `impliestan^(-1)x=(pi)/(6)` `impliesx=tan""(pi)/(6)=(1)/(sqrt3)` |
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| 132. |
यदि `sin^(-1)(1-x)-2sin^(-1)x=(pi)/(2)` तो x का मान बराबर है :A. `0,(1)/(2)`B. `1,(1)/(2)`C. 0D. `(1)/(2)` |
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Answer» Correct Answer - C `sin^(-1)(1-x)-2sin^(-1)x=(pi)/(2)` `{:(" माना "sin^(-1)x=theta),(implies" "x=sintheta):}` `impliessin^(-1)(1-sintheta)-2theta=(pi)/(2)` `impliessin^(-1)(1-sintheta)=((pi)/(2)+2theta)` `1-sintheta=sin((pi)/(2)+2theta)` `implies=cso2theta=1-2sin^(2)theta` `implies2sin^(2)theta-sintheta=0implies2x^(2)-x=0` `impliesx(2x-1)=0impliesx=0" या "x=(1)/(2)` परन्तु `x=(1)/(2)` से दिया समीकरण सन्तुष्ट नहीं होता। `:.x=0` |
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| 133. |
`sec^(2)(tan^(-1)4)+cosec^(2)(cot^(-1)3)` का मान है-A. 30B. 29C. 27D. 25 |
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Answer» Correct Answer - C |
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| 134. |
यदि `tan^(-1)x+tan^(-1)y=(pi)/(4),xy lt 1,` तब `x+y+xy` का मान ज्ञात कीजिए । |
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Answer» यहाँ `tan^(-1)x+tan^(-1)y=(pi)/(4)," यदि "xy lt 1` हम जानते हैं कि `tan^(-1)x+tan^(-1)y=tan^(-1)((x+y)/(1-xy)), xy lt 1` `therefore" "tan^(-1)((x+y)/(1-xy))=(pi)/(4)` `rArr" "(x+y)/(1-xy)= tan((pi)/(4))` `rArr" "(x+y)/(1-xy)=tan((pi)/(4))` `rArr" "(x+y)/(1-xy)=1` `rArr" "x+y=1-xy` `rArr" "x+y+xy=1.` |
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| 135. |
`2tan^(-1)""(1)/(2)+tan^(-1)""(1)/(7)=tan^(-1)""(31)/(17)` |
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Answer» `L.H.S.=2tan^(-1)""(1)/(2)+tan^(-1)""(1)/(7)` `=tan^(-1)((2xx(1)/(2))/(1-(1)/(4)))+tan^(-1)""(1)/(7)` `=tan^(-1)""(4)/(3)+tan^(-1)""(1)/(7)` `tan^(-1)""((4)/(3)+(1)/(7))/(1-(4)/(3)xx(1)/(7))=tan^(-1)""((28+3)/(21))/((21-4)/21)` `tan^(-1)""(31)/(17)=R.H.S.` |
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| 136. |
निम्नलिखित समीकरणों को हल कीजिए - `2tan^(-1)(cosx)=tan^(-1)("2 cosec x").` |
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Answer» दिया गया समीकरण है - `2tan^(-1)(cosx)=tan^(-1)("2 cosec x")` `rArr tan^(-1)((2cos x)/(1-cos^(2)x))=tan^(-1)((2)/(sinx))` `" "[because 2 tan^(-1)x=tan^(-1)((2x)/(1-x^(2)))]`. `rArr" "(2cosx)/(1-cos^(2)x)=(2)/(sinx)` `rArr" "(2cos x)/(sin^(2)x)=(2)/(sinx)` `rArr" "cos x=sinx` `rArr" "tanx=1` `rArr" "x=tan^(-1)(1)` `rArr" "x=tan^(-1)(tan.(pi)/(4))` `rArr" "x=(pi)/(4).` |
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| 137. |
मान निकालें : `sin^(-1)((sqrt(3))/(2))` |
| Answer» Correct Answer - `(pi)/(3)` | |
| 138. |
हल करें : `tan(cos^(-1)x) = sin(tan^(-1)2)` |
| Answer» Correct Answer - `pm(sqrt(5))/(3)` | |
| 139. |
सिद्ध कीजिए कि `tan^(-1)x+tan^(-1)""(2x)/(1-x^(2))=tan^(-1)""((3x-x^(3))/(1-3x^(2))),|x| lt (1)/(sqrt(3))` |
| Answer» Correct Answer - `((pi)/(4)+(x)/(2))` | |
| 140. |
सिद्ध कीजिए कि `sec^(2)(tan^(-1)3)+cosec^(2)(cot^(-1)2)=15.` |
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Answer» माना `tan^(-1)3=A" और "cot^(-1)2=B` `impliestanA=3" और "cotB=2` `L.H.S=sec^(2)(tan^(-1)3)+cosec^(2)(cot^(-1)2)` ltbgt `=sec^(2)A+cosec^(2)B=1+tan^(2)A+1+cot^(2)B` `=2+3^(2)+2^(2)=15=R.H.S` |
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| 141. |
सिद्ध कीजिए कि `cos^(-1)x=2sin^(-1)sqrt((1-x)/(2))` |
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Answer» माना `cos^(-1)x=thetaimpliesx=costheta` `R.H.S.=2sin^(-1)sqrt((1-x)/(2))=2sin^(-1)sqrt((1-costheta)/(2))` `=2sin^(-1)sqrt((2sin^(2)""(theta)/(2))/(2))=2sin^(-1)(sin""(theta)/(2))` `=2.(theta)/(2)=theta=cos^(-1)x=L.H.S.` |
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| 142. |
एक त्रिभुज के कोई दो कोण हो, तो तीसरा कोण `tan^(-1).(1)/(2)` एवं `tan^(-1).(1)/(3)` का मान ज्ञात कीजिए । |
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Answer» माना त्रिभुज का तीसरा कोण x है। दिया है - दो कोण, `y=tan^(-1).(1)/(2)" एवं "z=tan^(-1).(1)/(2)` `because` त्रिभुज के कोण नियम से, `x+y+z=pi` `rArr" "x+tan^(-1).(1)/(2)+tan^(-1).(1)/(3)=pi` `rArr" "x+tan^(-1)[((1)/(2)+(1)/(3))/(1-(1)/(2)xx(1)/(3))]=pi` `rArr" "x+tan^(-1)[(5//6)/(5//6)]=pi` `rArr" "x+tan^(-1)[1]=pi` `rArr" "x+(pi)/(4)=pi` `rArr" "X=pi-(pi)/(4)` `rArr" "x=(3pi)/(4)` |
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| 143. |
सिद्ध कीजिए कि - `cot^(-1)7+cot^(-1)8+cot^(-1)18=cot^(-1)3` |
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Answer» `L.H.S. = cot^(-1)7+cot^(-1)8+cot^(-1)18` `=tan^(-1).(1)/(7)+tan^(-1).(1)/(8)+tan^(-1).(1)/(18),` `" "[because cot^(-1)x=tan^(-1).(1)/(x)]` `=(tan^(-1).(1)/(7)+tan^(-1).(1)/(8))+tan^(-1).(1)/(18)` `=tan^(-1)(((1)/(7)+(1)/(8))/(1-(1)/(7)xx(1)/(8)))+tan^(-1).(1)/(18)` `=tan^(-1)((15)/(55))+tan^(-1).(1)/(18)` `=tan^(-1).(3)/(11)+tan^(-1).(1)/(18)` `=tan^(-1)((65)/(195))=tan^(-1)((1)/(3))` `=cot^(-1)3` = R.H.S. यही सिद्ध करना था। |
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| 144. |
सिद्ध कीजिए - `2tan^(-1)((1)/(2))+tan^(-1)((1)/(7))=tan^(-1)((31)/(17)).` |
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Answer» L.H.S. `=2tan^(-1)((1)/(2))+tan^(-1)((1)/(7))` `=tan^(-1).((2xx(1)/(2))/(1-((1)/(2))^(2)))+tan^(-1)((1)/(7)),` `[because 2tan^(-1)x=tan^(-1)((2x)/(1-x^(2)))]` `=tan^(-1)((4)/(3))+tan^(-1)((1)/(7))` `=tan^(-1)(((4)/(3)+(1)/(7))/(1-(4)/(3)xx(1)/(7)))` `=tan^(-1)(((28+3)/(21))/((21-4)/(21)))` `=tan^(-1).(31)/(17)` = R.H.S. |
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| 145. |
हल करें : `tan^(-1) ""(2alpha)/(1+alpha^(2)) + sin^(-1)""(2beta)/(1+beta^(2)) = 2tan^(-1)x,|alpha|le|1,|beta|le|1` |
| Answer» `(alpha+beta)/(1-alphabeta)` | |
| 146. |
सिद्ध कीजिए - `2tan^(-1)((1)/(5))+sec^(-1)((5sqrt2)/(7))+2tan^(-1).(1)/(8)=(pi)/(4)` |
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Answer» L.H.S. `=2tan^(-1)((1)/(5))+sec^(-1)((5sqrt2)/(7))+2tan^(-1)((1)/(8))` `=2[tan^(-1).(1)/(5)+tan^(-1).(1)/(8)]+tan^(-1)sqrt(((5sqrt2)/(7))^(3)-1),` `" "[because sec^(-1)x=tan^(-1)sqrt(x^(2)-1)]` `=2tan^(-1)(((1)/(5)+(1)/(8))/(1-(1)/(5)xx(1)/(8)))+tan^(-1)sqrt((50)/(49)-1),` `" "[because tan^(-1)x+tan^(-1)y=tan^(-1).(x+y)/(1-xy)]` `=2tan^(-1).(13)/(39)+tan^(-1).(1)/(7)` `=2tan^(-1).(1)/(3)+tan^(-1).(1)/(7)` `=tan^(-1)((2xx(1)/(3))/(1-((1)/(3))^(2)))+tan^(-1).(1)/(7),` `" "[because 2tan^(-1)x=tan^(-1)((2x)/(1-x^(2)))]` `=tan^(-1)(((2)/(3))/((8)/(9)))+tan^(-1).(1)/(7)` `=tan^(-1)((3)/(4))+tan^(-1).(1)/(7)` `=tan^(-1)(((3)/(4)+(1)/(7))/(1-(3)/(4)xx(1)/(7)))` `=tan^(-1)((25)/(25))` `=tan^(-1)(1)` `=tan^(-1)(tan.(pi)/(4))=(pi)/(4)" "[because tan^(-1)(Tan theta)=theta]` = R.H.S. यही सिद्ध करना था। |
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| 147. |
यदि `sin^(-1)x+sin^(-1)y=theta` हो तो सिद्ध कीजिए कि हम जानते हैं कि, - `cos^(-1)x+cos^(-1)y=pi-theta.` |
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Answer» हम जानते हैं कि, `sin^(-1)x+cos^(-1)x=(pi)/(2)` तथा`" "sin^(-1)y+cos^(-1)y=(pi)/(2)` समीकरण (1 ) और (2 ) को जोड़ने पर, `sin^(-1)x+sin^(-1)ycos^(-1)x+cos^(-1)y=(pi)/(2)+(pi)/(2)` दिया गया है - `sin^(-1)x+sin^(-1)y=theta` समीकरण (3 ) व (4 ) से, `theta+cos^(-1)x+cos^(-1)y=pi` `cos^(-1)x+cos^(-1)y=pi-theta`. यही सिद्ध करना था । |
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| 148. |
सिद्ध कीजिए कि `sin^(-1)""(3)/(5)=tan^(-1)""(3)/(4)` |
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Answer» माना `sin^(-1)""(3)/(5)=theta` `impliessintheta=(3)/(5)" implies "cosectheta=(5)/(3)` `impliescosec^(2)theta=(25)/(9)" implies "1+cot^(2)theta=(25)/(9)` `impliescot^(2)theta=(16)/(9)" implies "cottheta=(4)/(3)` `impliestantheta=(3)/(4)" implies "theta=tan^(-1)""(3)/(4)` `impliessin^(-1)""(3)/(5)=tan^(-1)""(3)/(4)` |
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| 149. |
सिद्ध कीजिए - `sin^(-1).(4)/(5)+sin^(-1).(5)/(13)+sin^(-1).(16)/(65)=(pi)/(2).` |
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Answer» L.H.S. `=sin^(-1).(4)/(5)+sin^(-1).(5)/(13)+sin^(-1).(16)/(65)` `=sin^(-1)[(4)/(5)xxsqrt(1-((5)/(13))^(2))+(5)/(13)xxsqrt(1-((4)/(5))^(2))]+sin^(-1).(16)/(65)` `=sin^(-1)[(4)/(5)xxsqrt((144)/(169))+(5)/(13)xx sqrt((9)/(25))]+sin^(-1).(16)/(65)` `=sin^(-1)((4)/(5)xx(12)/(13)+(5)/(13)xx(3)/(5))+sin^(-1).(16)/(65)` `=sin^(-1)((45)/(65)+(15)/(65))+sin^(-1).(16)/(65)` `=sin^(-1)((63)/(65))+sin^(-1)((16)/(65))` `=sin^(-1)[(63)/(65)sqrt(1-((16)/(65))^(2))+(16)/(65)sqrt(1-((63)/(65))^(2))]` `=sin^(-1)[(63)/(65)sqrt((4225-256)/(4225))+(16)/(65)xxsqrt((4225-3969)/(4225))]` `=sin^(-1)[(63)/(65)xxsqrt((3969)/(4225))+(16)/(65)xx sqrt((256)/(4225))]` `=sin^(-1)[(63)/(65)xx(63)/(65)+(16)/(65)xx(16)/(65)]` `=sin^(-1)((3969+256)/(4225))` `=sin^(-1)((4225)/(4225))` `=sin^(-1)=(pi)/(2)` = R.H.S. यही सिद्ध करना था। |
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| 150. |
सिद्ध कीजिए - `cot^(-1)((xy+1)/(x-y))+cot^(-1)((yz+1)/(y-z))+cot^(-1)((zx+1)/(z-x))=0,0 lt xy, yz, zx lt1.` |
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Answer» L.H.S. `cot^(-1)((xy+1)/(x-y))+cot^(-1)((yz+1)/(y-z))+cot^(-1)((zx+1)/(z-x))` `=tan^(-1)((x-y)/(1+xy))+tan^(-1)((y-z)/(1+yz))+tan^(-1)((z-x)/(1+zx))` `" "[because cot^(-1)x=tan^(-1).(1)/(x)]` `=(tan^(-1)x-tan^(-1)y)+(tan^(-1)y-tan^(-1)z)+(tan^(-1)z-tan^(-1)x),` `[because 0 lt xy, yz, zx lt 1]" और "[ tan^(-1)x-tan^(-1)y=tan^(-1)((x-y)/(1+xy))" यदि " xy gt -1]` = 0 = R.H.S. यही सिद्ध करना था। |
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