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6651.

Calculus

Answer»
6652.

Can you plz understand me suplimentry exercise ch i

Answer»
6653.

Prove that sin^2A+sin^2(60+A)+sin^2(60-A)=3/2

Answer» {tex}\\begin{array}{l}LHS=\\sin^2A+\\sin^2(60^\\circ+A)+\\sin^2(60^\\circ-A)\\\\=\\sin^2A+\\frac12\\left[1-\\cos2(60^\\circ+A)\\right]+\\frac12\\left[1-\\cos2(60^\\circ-A)\\right]\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\lbrack\\sin ce\\;\\;\\sin^2\\theta=\\frac12(1-\\cos2\\theta)\\;\\;\\rbrack\\\\=\\sin^2A+1-\\frac12\\left[\\cos(120^\\circ+2A)+\\cos(120^\\circ-2A)\\right]\\\\=\\sin^2A+1-\\frac12\\times2\\cos\\frac{(120^\\circ+2A)+(120^\\circ-2A)}2.\\cos\\frac{(120^\\circ+2A)-(120^\\circ-2A)}2\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\sin ce\\;\\;\\cos\\;C\\;+\\;\\cos\\;D=2.\\cos\\frac{C+D}2.\\cos\\frac{C-D}2\\\\=\\sin^2A+1-\\cos120^\\textdegree.\\cos2A\\\\=\\sin^2A+1+\\frac12.\\left(1-2\\sin^2A\\right)\\\\=\\sin^2A+1+\\frac12-\\sin^2A\\\\=\\frac32=RHS\\end{array}{/tex}
6654.

what is the max. value of sinx cosx

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6655.

Solution of tan -pie/18

Answer»
6656.

Cos20.cos40.cos60.cos80

Answer» It is there in Mathematic
6657.

Why cosec is reciprocal of sine

Answer»
6658.

2cos +sin4

Answer»
6659.

What is a domain and range of functions sin cos tan cot sec cosec

Answer» Sin: Domain:- RRange:- [-1,1]Cos:Domain:- RRange:- [-1,1]Tan:Domain:- R-{(2n+1)pie/2}Range:-RCosec:Domain:- R-{n×pie}Range:- R-(-1,1)Sec:Domain:- R-{(2n-1)pie/2}Range:- R-(-1,1)Cot:Domain:- R-{n×pie}Range:- R
6660.

Sin c+ sin d =

Answer» 2sin c+d/2.cos c-d/2
6661.

Limit x-0 sin³x/x Limit x-0 sinx³/x

Answer»
6662.

Last year questions paper

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6663.

if f(x) = 9^x upon 9^x+3 then f (x) + f ( 1-x)

Answer» 1
6664.

24 +32

Answer»
6665.

7+77+777+7777+77777+---------+n. Solv3 the series

Answer»
6666.

Values of trigonometric ratios

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6667.

Complex numbers

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6668.

Jsksnskx

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6669.

Find the square root of the complex numbers:- 5 - 12 iota

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6670.

Sin,255

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6671.

4X2 +3X+3=

Answer» Not understand questions
6672.

{tex}[i^{18}+(1/i)^{25}]^3{/tex}

Answer» ={(i^2)^9-[(i)^2]^12.i}^3=[-i-(-1)^12.i]^3=(-i-i)^3=(-2i)^3=-8(i^3)=8i
6673.

2nç3:ñç3=12:1

Answer» n=5 so u can solve it now
6674.

COSX+COSX=COS2X+COS4X

Answer»
6675.

What is DE- MORGAN\'S LAWS ( Sets)

Answer» (A intersection B)\'=A\'UB\'
6676.

1212+562

Answer» 1774
6677.

How many no\'s. Greater than 20000 can be formed by using 0,1,2,3,4 with no repetition allowed

Answer» 72
6678.

Modulus

Answer» √asquare+bsquare
6679.

if a(1/b+1/c),b(1/a+1/c),c(1/a+1/b) are in A.p,prove that a,b,c are in a.p

Answer»
6680.

Write down domain & range 1+2/1+3

Answer» Wrong question
6681.

3*5/8+6-2=?

Answer» 5.8(approx)
6682.

In triangle ABC ,if c=3.4cm ,A=25° ,B=85°, find a,b and angle C

Answer»
6683.

Tan15 degree + cos 15 degree=4

Answer» Tan15°+cos15°=2root2-root3+2+root3/2root2+root3
6684.

Cos 1 degree. Cos 2degree.................. Cos 100degree

Answer» The answer is zero. Since cos 1. Cos2...... Cos90...... Cos 100. = 0
6685.

tan15 degree + cos 15 degree =4

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6686.

If tan A × tan B = root under a-b/a+b then prove that (a-b cos 2A) × (a-b cos 2B)= a×a-b×b

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6687.

Why rectangle have 360 degree sum of angle

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6688.

prove that the centre of three circles are collinear

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6689.

Find the value of tanπ÷8

Answer» Let twice th e angle so 2×tanπ\\8=1so 2t/1-tsquare,=√2-1
6690.

Rhg

Answer»
6691.

2n

Answer» If it is question of induction then for n=m 2m <(m+2) for n=m+1, 2 (m+1)=2m+1, <(m+2)+1 <(m+1)+2 which gives 2 (m+1)<(m+1+2) by pmi it is true
6692.

By mathematical induction show that n(nsquare +11) is divisble by 6

Answer»
6693.

434+432

Answer» 866
6694.

a=cos¥+isin¥,find the value of 1+a/1-a

Answer» Bxjwjckde
6695.

Sin A + cosB _cos C

Answer»
6696.

Value of sin22.5°

Answer» According to the rational root theorem, the possible rational roots are ±1, ±2, ±4, and ±8.5)To find the exact value of sin(22.5 degrees), we write this as a half angle.sin(45 / 2)
6697.

Limits

Answer»
6698.

Cos4x = cos2x

Answer» Cos2(2X )=cos2xPut the formula is cod 2x On LHS side LEt 2x=x
6699.

What is value of sine 10°

Answer»
6700.

Complex numbers exercise miscellaneous q.no.17

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