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Assume that log 2 is rational, that is,

log2=p/q(1)

where p, q are integers.

Since log 1 = 0 and log 10 = 1,0 < log 2 < 1and therefore p < q.

From (1),2= 10^(p/q)2^q=(2×5)^p2(p-q)=5^p

where q – p is an integer greater than 0.

Now, it can be seen that the L.H.S. is even and the R.H.S. is odd.

Hence there is contradiction andlog 2is irrational.

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Sin30°=1/2

sin90°=1

cos0°=1

tan30°=1√3

tan60°=√3

now putting these hese value on question

so..we find

1/2-1+2×1/1/√3 ×√3

1/2-1+2/√3/√3

1-2+4/2

3/2 ANSWER

8*15 = x*10

x = 8*15/10

x = 8*3/2

x = 12 pages

Tan²θ=1-a²LHS=secθ+tan³θcosecθ=√(1+tan²θ)+tan²θ×tanθ×√(1+cot²θ)[∵, sec²θ-tan²θ=1 and cosec²θ-cot²θ=1]=√1+(1-a²)+(1-a²)×√(1-a²)×√{1+(1/tan²θ)}=√(2-a²)+(1-a²)×√(1-a²)×√{1+1/(1-a²)}=√(2-a²)+(1-a²)×√(1-a²)×√{(1-a²+1)/(1-a²)}=√(2-a²)+(1-a²)×√(2-a²)=√(2-a²)×(1+1-a²)=√(2-a²)×(2-a²)=(2-a²)¹/²⁺¹=(2-a²)³/²=RHS (Proved)

1. is the best answer

it is formula of tan (a+b)

tan (30+60) calculate

sorry do not seen the negatives sign upon

sorry not seen upon sign

It is expanison Tan(a-b) = Tan(60 - 30) = Tan 30 = 1/root3

1 is the correct answer

1/root3 is the correct answer

1-√3is right answer........

1/ root 3 is the correct answer

,2/root3 is the best answer me

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sin60°=√3/2R. H. S 2/√3/1+1/3= √3/2

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it's a GP seriesas comman ratio is same

as log base 6 (2) / log base 3(2)= 2

And log base (12) 2/ log base(6) 2 = 2 as well

2x-3y=18let X be 0hence y=-6so y intercept is (0,-6)let's assume y=0hencex=9hence X intercept is (0,9)

it's answer is 1 ratio 9

1 and 3 is right .........

1 and 3 is right way of this question

1 and 3 is right way of this question

tanx^3-1/tanx-1=tanx^2-1==tanx-(tanx^2-secx^2)=tanx-tanx^2-secx^2= tanx+secx^2

Given,Sec theta = 2Sec theta = Sec 60°

theta = 60°

Therefore,4cos theta - root(3).sin theta/tan theta - cot theta

= 4 cos 60 - root(3).sin 60/ tan 60 - cot 60

= 4*1/2 - root(3)*root(3)/2 / root(3) - 1/root(3)

= (4 - 3)/2 / (3 - 1)/root(3)

= 1/2 / 2/root(3)

= root(3)/4

4000

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(4×4×4×4×4×4)/(4×4×4×4×4×4×4×4×4) = 1/(4×4×4) = 1/64

Ratio of 500cm to 10 cm is

500 cm----------10 cm

50/1

So, ratio is 50 : 1

(3x+ 8) (5 - 2x)15x - 6x² + 40 -16x-6x² - x + 40

thank you mam

LHS = (sinθ - cosθ + 1)/(sinθ + cosθ - 1) dividing by cosθ both Numerator and denominator = (sinθ/cosθ - cosθ/cosθ + 1/cosθ)/(sinθ/cosθ + cosθ/cosθ - 1/cosθ)= (tanθ + secθ - 1)/(tanθ - secθ + 1)

Multiply (tanθ - secθ) with both Numerator and denominator = (tanθ + secθ - 1)(tanθ - secθ)/(tanθ - secθ + 1)(tanθ - secθ)= {(tan²θ - sec²θ) - (tanθ - secθ)}/(tanθ - secθ + 1)(tanθ - secθ)= (-1 - tanθ + secθ)/(tanθ - secθ + 1)(tanθ - secθ) [ ∵ sec²x - tan²x = 1 ]= -1/(tanθ - secθ) = 1/(secθ - tanθ) = RHS

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18over5,1 2over5 answer of 1st sum

(1) 3 3/5 ÷ 2 2/5= 18/5 ÷ 12/5= 18/5 * 5/12= 3/2

(2) 1 7/104 ÷ 11/13= 111/104 ÷ 11/13= 111/104 * 13/11= 111/8*11= 111/88

(1) 33/5÷22/5= 18/5÷ 12/5= 18/5*5;12= 3/2

as a remainder is zero and the quotient is non zero

x 4/3 ÷ x 3/4; 4/3 × 3/4=16+9/12=25/12