InterviewSolution
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InterviewSolution
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.
| 2751. |
Value of tan10 |
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| 2752. |
If sinA +cosA=x then show that sin^6A+cos^6A=4-3(x^2-1)^2/3 |
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| 2753. |
Hey ANJU |
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| 2754. |
Ramshi Aditi r u online |
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| 2755. |
Aditi sirf 2 min ke liya baat karlo please |
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| 2756. |
Aditi message me?? |
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| 2757. |
Aditi message toh karo whatspp pe jaldi |
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| 2758. |
Prove that cot a_cos a |
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| 2759. |
Sbke sb so gye okk byy ANJU gud ni8 |
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| 2760. |
Is there anyone for time passing plzz |
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| 2761. |
Hey gud ni8 gyzzz |
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| 2762. |
Sahil me tumhara behana nhi hu |
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| 2763. |
Ramshi where r u |
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| 2764. |
Area of a Square |
| Answer» Side square | |
| 2765. |
empirical formula |
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| 2766. |
Formuale of segment |
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| 2767. |
Draw the less than ogive and more than ogive |
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| 2768. |
2sec2a-sec4a+2cosec2a+cosec4a=cot4a-tan4a |
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| 2769. |
128x24 |
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Answer» 3072 3072 |
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| 2770. |
Is it important to slove the additional exercises |
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| 2771. |
Cylindr |
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| 2772. |
Concide point |
| Answer» Two or more points on a point | |
| 2773. |
Solve the following system of linear equations by using the method of elimination 2x-3y=7 and x+y=1? |
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| 2774. |
What is ur dream |
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| 2775. |
when is tenth cbse board final exam going to start ? |
| Answer» March | |
| 2776. |
(CosecA-sinA) (secA-cosA) (tanA+cotA) =1 |
| Answer» To prove:(cosecA - sinA) (secA - cosA) (tanA + cotA) = 1LHS\xa0{tex}= (\\cos ecA - \\sin A)(\\sec A - \\cos A)(\\tan A + \\cot A){/tex}{tex} = \\left( {\\frac{1}{{\\sin A}} - \\sin A} \\right)\\left( {\\frac{1}{{\\cos A}} - \\cos A} \\right)\\left( {\\frac{{\\sin A}}{{\\cos A}} + \\frac{{\\cos A}}{{\\sin A}}} \\right){/tex}\xa0{tex}\\left[ \\begin{gathered} \\because \\cos ecA = \\frac{1}{{\\sin A}},\\sec A = \\frac{1}{{\\cos A}}, \\hfill \\\\ \\tan A = \\frac{{\\sin A}}{{\\cos A}},\\cot A = \\frac{{\\cos A}}{{\\sin A}} \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\left( {\\frac{{1 - {{\\sin }^2}A}}{{\\sin A}}} \\right)\\left( {\\frac{{1 - {{\\cos }^2}A}}{{\\cos A}}} \\right)\\left( {\\frac{{{{\\sin }^2}A + {{\\cos }^2}A}}{{\\cos A\\sin A}}} \\right){/tex}{tex} = \\frac{{{{\\cos }^2}A}}{{\\sin A}} \\times \\frac{{{{\\sin }^2}A}}{{\\cos A}} \\times \\frac{1}{{\\cos A\\sin A}}{/tex}\xa0{tex}\\left[ \\begin{gathered} \\because 1 - {\\sin ^2}A = {\\cos ^2}A,1 - {\\cos ^2}A = {\\sin ^2}A, \\hfill \\\\ {\\sin ^2}A + {\\cos ^2}A = 1 \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\frac{{\\cos A \\times \\cos A}}{{\\sin A}} \\times \\frac{{\\sin A \\times \\sin A}}{{\\cos A}} \\times \\frac{1}{{\\cos A\\sin A}}{/tex}= 1= RHS | |
| 2777. |
Show that any positive odd integer is of the form 6q+1,or6q+3,or6q+5, where q is some integer |
| Answer» Let a be any positive integer and b = 6∴ by Euclid’s division lemmaa = bq + r, 0≤ r and q be any integer q ≥ 0∴ a = 6q + r,where, r = 0, 1, 2, 3, 4, 5If a is even then then remainder by division of 6 is 0,2 or 4Hence r=0,2,or 4or A is of form 6q,6q+2,6q+4As, a = 6q = 2(3q), ora = 6q + 2 = 2(3q + 1), ora = 6q + 4 = 2(3q + 2).If these 3 cases a is an even integer.but if the remainder is 1,3 or 5 then r=1,3 or 5or A is of form 6q+1,,6q+3 or,6q+5Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,\xa0Case 2: a = 6q + 3 = 6q + 2 + 1,= 2(3q + 1) + 1 = 2n + 1,\xa0Case 3: a = 6q + 5 = 6q + 4 + 1= 2(3q + 2) + 1 = 2n + 1This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. | |
| 2778. |
Some important question ncert chapter 1 |
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| 2779. |
Show that there is no positive integer n for which √n-1 +√n+1 is rational. |
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| 2780. |
Cot A-cosA/cotA+cosA=cosecA-1/cosecA+1 |
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| 2781. |
Quize |
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| 2782. |
How to download sample paper |
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| 2783. |
Formula of rhombus |
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Answer» 1/2* products of its diagonals (1/2 )×diagonal 1× diagonal 2 |
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| 2784. |
Hwy RIMSHA are u here |
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| 2785. |
Okay I\'m going byyyy |
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| 2786. |
Where r u RIMSHA |
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| 2787. |
RIMSHA come plzzz |
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| 2788. |
What is the ratio of volume of cube to that of the sphere which will fit inside it |
| Answer» Let the radius of the sphere which fits exactly into a cube be r units. The length of each edge of cube = 2r unitsLet V1 and V 2 be the volumes of the cube and sphereThen V1 = (2r)3{tex}{V_2} = \\frac{4}{3}\\pi {r^3}{/tex}{tex}\\frac{{{V_1}}}{{{V_2}}} = \\frac{{8{r^3}}}{{\\frac{4}{3}\\pi {r^3}}} = \\frac{6}{\\pi }{/tex}V1:V2\xa0{tex} = 6:\\pi {/tex} | |
| 2789. |
335 and 225 real number |
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| 2790. |
135 and 225 |
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| 2791. |
Hey eshaa talk to me |
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| 2792. |
if pth, qth,& rth term of an AP are a,b,c respectively, then show that a(q-r)+b(r-p)+c(p-q) |
| Answer» Let the first term and common difference of the AP be A and Dpth term = a ……… (given)= A + (p – 1)D = a\xa0............(1)qth term = b{tex}\\Rightarrow {/tex}\xa0A + (q - 1) D = b ......... (2)rth term = c .......... Given{tex}\\Rightarrow {/tex}\xa0A + (r - 1) D = c ........ (3)Multiplying equations (1), (2) and (3) by q - r, r - p and p - q respecitvely, we geta(q - r) + b(r - p) + c(p - q)= [A + (p - 1) D] (q - r) + [A + (q - 1)D] (r - p) + [A + (r - 1) D] (p - q)]= A [q - r + r - p + p - q] + D [(p - q)\xa0(q - r) + (q\xa0- 1) (r- p) + (r - 1) (p - q)]= A (0) + D(0)= 0 | |
| 2793. |
Find the area of sector of a circle of radius R /2 and angle subtended at the centre is 2p |
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| 2794. |
If sinA +sinB + sinC=3 then find the value of cosA+cosB+cosC=? |
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| 2795. |
What happened RIMSHA |
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| 2796. |
4x2_4A2X _ A4_B4 |
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| 2797. |
CosA-sinA=1,prove sinA+cosA=+-1 |
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| 2798. |
RIMSHA plzzz |
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| 2799. |
CotA+cosecA-1\\cotA-coseA+1=1+cosA\\1+sinA |
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| 2800. |
How many of you using fb???? |
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