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  • 1 answers

Äðřîjã Păĺ 6 years, 4 months ago

x = cot(Π/6) = √3
  • 3 answers

Honey Rajput 6 years, 4 months ago

-pie by 4 is correct

Honey Rajput 6 years, 4 months ago

Its pie by 4

Äðřîjã Păĺ 6 years, 4 months ago

15Π/4
  • 5 answers

Deepanshi Garg 6 years, 4 months ago

Syllabus change as there are 20 mcq of 1 marks and remaining as usual

♥️Bhavika Naik?? 6 years, 4 months ago

Ha...we will 30 marks internal & in boards 20 ques will b MCQs n remaining 50 mrks will b written part

Aman Kumar 6 years, 4 months ago

This year, 20 questions of 1 marks( form of MCQ )

Äðřîjã Păĺ 6 years, 4 months ago

Refer to the syllabus 2019-20

Vikram Kumar 6 years, 4 months ago

No change anything
  • 2 answers

Anisha Sharma 6 years, 4 months ago

I think my way of studying maths is not right...... That's why I'm not getting the concepts clearly..... Please suggest best way to study maths.... How to manage tuitions study and self study.... Sometimes I feel like I am so stupid and dumb..... Till 10 I was good in maths studying without any tuitions but from 11 even after taking maths tuition... Pehle se bht bekaar halat Ho gyi h..... Kaise thik kru maths ki problem? ??

Aman Kumar 6 years, 4 months ago

If your basics is not getting clear and could not able to solve the question ,then according to me.....whenever u study any properties of derivatives or any methods....just go for some problems based on it, practice it and when u satisfied that u had a clear concept of it, then go to next one...do the same thing for all the properties and methods........in this way...u may get this chapter easy..
  • 1 answers

Varun Sharma 6 years, 4 months ago

10x4 sinx5 cosx5 cosx3-3x2 sinx3 sin2x5
  • 0 answers
  • 4 answers

Karan Gr 6 years, 4 months ago

Same problem yr

Dolly ?️ 6 years, 4 months ago

Bhi ya kam karo; 1)read all formulas. 2)give atleast 2 days for this chapter only. Aur aur Do Din Mein only and only inverse trigo k question karo aur. if formula Yaad nahi hote Tu question karte samay informulo co dekh kar questions Caro Jitna ho sake utnai Jyada questions karo.

Aditya Yadav 6 years, 4 months ago

a b c a-b b-c c-a = a³+b³+c³-3abc b+c c+a a+b

Sia ? 6 years, 4 months ago

You should do following things :
Make sure you know all the formulas. They are the base of this chapter. Don't do rote learning. Rather memorize by solving problems.
Solve NCERT thoroughly.
Solve all the questions of NCERT Exempler, specially the subjective ones.
Solve previous year's questions. It will help you to get acclimatized to the type of question asked and also help you in filling loopholes in your preparation.
Inverse Trigonometric Function is a scoring chapter. All it require is good grasp of formula along with adequate practice. Make sure you don’t lose easy marks which you can get from this chapter.
  • 2 answers

Nishu Singh 6 years, 4 months ago

If it was sin-¹(sin1550) Then we know that sin inverse of sin x = x so sin inverse sin 1550 = 8.611π

Nishu Singh 6 years, 4 months ago

sin-¹(1800-250) sin-¹250 sin-¹(180+70) -sin-¹(70) -0.939
  • 1 answers

Sia ? 6 years, 4 months ago

Here, we have to prove that {tex} \cos \left[\sin ^ { - 1 }( \frac { 3 } { 5 } )+ \cot ^ { - 1 }( \frac { 3 } { 2 } \right) ]= \frac { 6 } { 5 \sqrt { 13 } }{/tex}
Let  us consider , {tex} \sin ^ { - 1 } (\frac { 3 } { 5 }) = x{/tex} and {tex} \cot ^ { - 1 } (\frac { 3 } { 2 }) = y ; \forall x \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]{/tex}
and {tex} y \in ( 0 , \pi ){/tex}
{tex} \Rightarrow \quad \sin x = \frac { 3 } { 5 }{/tex}and {tex}\cot y = \frac { 3 } { 2 }{/tex}
{tex}\Rightarrow \quad \cos x = \sqrt { 1 - \sin ^ { 2 } x }{/tex}and 

cosec y ={tex}\sqrt { 1 + \cot ^ { 2 } y }{/tex}{tex}\left[ \begin{array} { l } { \text { taking positive sign as } x \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right] } \\ { \text { and } y \in ( 0 , \pi ) } \end{array} \right]{/tex}
{tex}\Rightarrow \cos x = \sqrt { 1 - \left( \frac { 3 } { 5 } \right) ^ { 2 } }{/tex}and cosec y {tex}= \sqrt { 1 + \left( \frac { 3 } { 2 } \right) ^ { 2 } }{/tex}
{tex}\Rightarrow \quad \cos x = \sqrt { 1 - \frac { 9 } { 25 } }{/tex}and cosec y {tex} = \sqrt { 1 + \frac { 9 } { 4 } }{/tex}
{tex}\Rightarrow \quad \cos x = \sqrt { \frac { 25 - 9 } { 25 } } = \sqrt { \frac { 16 } { 25 } } = \frac { 4 } { 5 }{/tex}
and cosec y {tex} = \sqrt { \frac { 4 + 9 } { 4 } } = \sqrt { \frac { 13 } { 4 } } = \frac { \sqrt { 13 } } { 2 }{/tex}
{tex}\Rightarrow \quad \cos x = \frac { 4 } { 5 }{/tex} and {tex}\frac { 1 } { \sin y } = \frac { \sqrt { 13 } } { 2 }{/tex}
{tex}\Rightarrow \quad \cos x = \frac { 4 } { 5 }{/tex} and {tex}\sin y = \frac { 2 } { \sqrt { 13 } }{/tex}
Also, cosy = siny - coty = {tex} \frac { 2 } { \sqrt { 13 } } \times \frac { 3 } { 2 } = \frac { 3 } { \sqrt { 13 } }{/tex}
Now, cos(x + y) = cosx cosy - sinx siny {tex} = \frac { 4 } { 5 } \times \frac { 3 } { \sqrt { 13 } } - \frac { 3 } { 5 } \times \frac { 2 } { \sqrt { 13 } }{/tex}{tex} = \frac { 12 } { 5 \sqrt { 13 } } - \frac { 6 } { 5 \sqrt { 13 } } = \frac { 6 } { 5 \sqrt { 13 } } = \mathrm { RHS }{/tex} 
Hence proved.

  • 5 answers

Deepanshi Garg 6 years, 4 months ago

ex(sinx-cosx)/sin^2x answer

Dolly ?️ 6 years, 4 months ago

(e^x(sinx) - cosx(e^x)) /sin^2 x e^x(sinx-cosx) /sin^2 x e^x. Cosecx- e^x. cotx. Cosecx e^xcoescx(1-cotx)

Santy Baralu 6 years, 4 months ago

Apply formula of dv /dx. You got the answer

Santy Baralu 6 years, 4 months ago

e*cosecx (1- cotx)

Nancy Rajput 6 years, 4 months ago

e^x(cosx+sinx)
  • 2 answers

Deepanshi Garg 6 years, 4 months ago

Let y = cosx.cos2x.cos3x Taking log on both sides,we get Logy=log(cosx)+log(cos2x)+log(cos3x) Differentiation with respect to 'x' d/dx(logy)=d/dx(log(cosx))+d/dx(log(cos2x))+d/dx(log(cos3x)) 1/y.dy/dx=1/cosx(-sinx)+1/cos2x(-sin2x).2+1/cos3x(-sin3x).3 dy/dx= -y[tanx+2tan2x+3tan3x] dy/dx= -cosx.cos2x.cos3x[tanx+2tan2x+3tan3x] Answer

Deepak Singh Rautela 6 years, 4 months ago

You can solve this by taking log .
  • 0 answers
  • 1 answers

Sia ? 6 years, 4 months ago

Differentiation allows us to find rates of change.
  • 1 answers

Sia ? 6 years, 4 months ago

Let, cos\(^{-1}\) x = α and cos\(^{-1}\) y = β

From cos\(^{-1}\) x = α we get,

x = cos α

and from cos\(^{-1}\) y = β we get,

y = cos β

Now, cos (α - β) = cos α cos β + sin α sin β

⇒ cos (α - β) = cos α cos β + \(\sqrt{1 - cos^{2} α}\) \(\sqrt{1 - cos^{2} β}\)

⇒ cos (α - β) = (xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

⇒ α - β = cos\(^{-1}\)(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

or, cos\(^{-1}\) x - cos\(^{-1}\) y = cos\(^{-1}\)(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

Therefore, arccos(x) - arccos(y) = arccos(xy) + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))       Proved.

 

Note: If x > 0, y > 0 and x\(^{2}\) + y\(^{2}\) > 1, then the cos\(^{-1}\) x + sin\(^{-1}\) y may be an angle more than π/2 while cos\(^{-1}\)(xy - \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\)), is an angle between – π/2 and π/2.

Therefore, cos\(^{-1}\) x - cos\(^{-1}\) y = π - cos\(^{-1}\)(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

  • 2 answers

Amrita Singh 6 years, 4 months ago

π/2

Sia ? 6 years, 4 months ago

Let, sin\(^{-1}\) x = α and sin\(^{-1}\) y = β

From sin\(^{-1}\) x = α we get,

x = sin α

and from sin\(^{-1}\) y = β we get,

y = sin β

sin (α + β) = sin α \(\sqrt{1 - sin^{2} β}\) + \(\sqrt{1 - sin^{2} α}\) sin β

sin (α + β) = x ∙ \(\sqrt{1 - y^{2}}\) + \(\sqrt{1 - x^{2}}\) ∙ y

Therefore, α + β = sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\)) 

or, sin\(^{-1}\) x + sin\(^{-1}\) y = sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\)).       Proved.


Note: If x > 0, y > 0 and x\(^{2}\) + y\(^{2}\) > 1, then the sin\(^{-1}\) x + sin\(^{-1}\) y may be an angle more than π/2 while sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\)), is an angle between – π/2 and π/2.

Therefore, sin\(^{-1}\) x + sin\(^{-1}\) y = π - sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))



2a
  • 1 answers

Mukesh Jani 6 years, 4 months ago

Is this a complete question
  • 1 answers

Sia ? 6 years, 4 months ago

P (RR)+P (BB)+P (GG)={tex}\frac{3 \times 2 + 4 \times 3 + 2 \times 1}{9 \times 8}{/tex} = {tex}\frac{20}{72} = \frac{5}{18}{/tex}
Since this is the probability of a match, the probability that we do not match colors is {tex}1 - \frac{5}{18} = \frac{13}{18}{/tex}

  • 1 answers

Sandeep Yadav 6 years, 4 months ago

Good evening sir
  • 2 answers

Pankaj Chadda 6 years, 4 months ago

X is angle

Hardik Singh 6 years, 4 months ago

Its just a function u have to differentiate it Cosx.cosx will ne the answer
  • 2 answers

Jabra Fan ? 6 years, 4 months ago

Thnx bro

Äðřîjã Păĺ 6 years, 4 months ago

It is the result of multiplying a given number of consecutive integers from 1 to the given number. In equations, it is symbolized by an exclamation mark (!). For example, 5! = 1 × 2 × 3 × 4 × 5 = 120.
  • 3 answers

Aman Yadav 6 years, 4 months ago

5 min.

Jabra Fan ? 6 years, 4 months ago

Done

Nikkie Gupta 6 years, 4 months ago

Plzz prove it

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