Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Ayush Binjola 6 years, 4 months ago
- 1 answers
Posted by Ayush Binjola 6 years, 4 months ago
- 1 answers
Posted by Ayush Binjola 6 years, 4 months ago
- 3 answers
Posted by Ayush Binjola 6 years, 4 months ago
- 0 answers
Posted by Dikshant Bhanot 6 years, 4 months ago
- 5 answers
Deepanshi Garg 6 years, 4 months ago
♥️Bhavika Naik?? 6 years, 4 months ago
Posted by Anisha Sharma 6 years, 4 months ago
- 2 answers
Anisha Sharma 6 years, 4 months ago
Aman Kumar 6 years, 4 months ago
Posted by Harshit Kumar 6 years, 4 months ago
- 0 answers
Posted by Varun Sharma 6 years, 4 months ago
- 1 answers
Posted by Aakashdeep Singh Chandi 6 years, 4 months ago
- 0 answers
Posted by Anjali Thakur 6 years, 4 months ago
- 0 answers
Posted by Sweta Nath 6 years, 4 months ago
- 4 answers
Dolly ?️ 6 years, 4 months ago
Sia ? 6 years, 4 months ago
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.
Posted by Sonu Redhu Jaat Pardan 6 years, 4 months ago
- 1 answers
Posted by Ayush Binjola 6 years, 4 months ago
- 2 answers
Nishu Singh 6 years, 4 months ago
Nishu Singh 6 years, 4 months ago
Posted by Asif Parwez 6 years, 4 months ago
- 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.
Posted by Himanshu Arya 6 years, 4 months ago
- 5 answers
Dolly ?️ 6 years, 4 months ago
Posted by Himanshu Arya 6 years, 4 months ago
- 2 answers
Deepanshi Garg 6 years, 4 months ago
Posted by Mohit Kumar 6 years, 4 months ago
- 0 answers
Posted by Abhi Sindriya 6 years, 4 months ago
- 0 answers
Posted by Abhi Sindriya 6 years, 4 months ago
- 1 answers
Posted by Aman Singh 6 years, 4 months ago
- 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 + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))
Posted by Aman Singh 6 years, 4 months ago
- 2 answers
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\(\sqrt{1 - x^{2}}\))
Posted by Yashu Mehra 6 years, 4 months ago
- 1 answers
Posted by Shivani Ramu 6 years, 4 months ago
- 0 answers
Posted by Mehak Thakur 6 years, 4 months ago
- 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}
Posted by Parteek Gurjar 6 years, 4 months ago
- 1 answers
Posted by Ramen Mahapatra 6 years, 4 months ago
- 2 answers
Hardik Singh 6 years, 4 months ago
Posted by Jabra Fan ? 6 years, 4 months ago
- 2 answers
Äðřîjã Păĺ 6 years, 4 months ago
Posted by Nikkie Gupta 6 years, 4 months ago
- 3 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Äðřîjã Păĺ 6 years, 4 months ago
0Thank You