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PSGR Krishnammal College For Women Humanities & Social Sciences

30. cos²2x – cos²6x = sin 4x sin 8x

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Payal Gupta

Contributor-Level 10

30. L.H.S $L.H.S = c o s^{2} 2 x - c o s^{2} 6 x$

$= (c o s 2 x + c o s 6 x) (c o s 2 x - c o s 6 x) [? a^{2} - b^{2} = (a - b) (a + b)]$

L.H.S =

L.H.S = [2 c o s (\frac{2 x + 6 x}{2}) c o s (\frac{2 x - 6 x}{2})] [- 2 s i n \frac{2 x + 6 x}{2} s i n (\frac{2 x - 6 x}{2})]

= [2 c o s \frac{8 x}{2} c o s (- \frac{4 x}{2})] [- 2 s i n (\frac{8 x}{2}) s i n (- \frac{4 x}{2})]

= 2 c o s 4 x c o s 2 x \times 2 s i n 4 x \cdot s i n 2 x [? c o s (- x) = c o s x; s i n (- x) = - s i n x]

= 2 s i n 4 x c o s 4 x \times 2 s i n 2 x c o s 2 x

As sin 2θ = 2 sinθ cosθ.

L.H.S. = sin (2*4x) sin(2*2x)

= sin 8x sin 4x

= R.H.S.

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10 months ago

I want to know about the latest cutoff information for B.A. in History at PSGR Krishnammal College for Women.

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Chandra Kumar

Contributor-Level 6

The latest cutoff marks for B.A. in History at PSGR Krishnammal College for Women vary depending on the category (general, OBC, etc.) and yearthe CUET UG entrance exam percentile is a factor in determining the cutoff, with a CUET cut-off of 100 for the general category and 99 for the OBC category. Additionally, applicants must have secured 40% marks in the aggregate and 50% marks in the subject concerned. To get the most up-to-date information, it's best to check the official PSGR Krishnammal College website for the latest admission announcements.

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10 months ago

44. Find the equation of the plane passing through $(a, b, c)$ and parallel to the plane $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 2$

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Balaji Institute of Technology Business Management Studies

Contributor-Level 10

Any plane parallel to the plane, $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 2$ , is of the form $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) = λ . . . . . . . . . (1)$

The plane passes through the point (a, b, c). Therefore, the position vector $\vec{r}$ of this point is $\vec{r} = a \hat{i} + b \hat{j} + c \hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l} (a \hat{i} + b \hat{j} + c \hat{k}) . (\hat{i} + \hat{j} + \hat{k}) = λ \\ \Rightarrow a + b + c = λ \end{array}$

Substituting $λ = a + b + c$ in equation (1), we obtain

$\vec{r} = (\hat{i} + \hat{j} + \hat{k}) = a + b + c . . . . . . . . . (2)$

This is the vector equation of the required plane.

Substituting $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ in equation (2), we obtain

$\begin{array}{l} (x \hat{i} + y \hat{j} + z \hat{k}) . (\hat{i} + \hat{j} + \hat{k}) = a + b + c \\ \Rightarrow x + y + z = a + b + c \end{array}$

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10 months ago

What are the benefits of pursuing MBA from Balaji Institute of Technology Management?

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Parul Shukla

Contributor-Level 10

The MBA eligibility requires candidates to have completed their graduation in any field (Science, Commerce, Arts) with an aggregate of 50% or more. Students can also check the list of reasons below to pursue MBA course from Balaji Institute of Technology Management:

Promising Career: The simplest approach to advance your career is to earn a Master of Business Administration. Students concentrate on the employment possibilities that follow earning a Master of Business Administration. MBA courses are becoming increasingly important to have managerial abilities and expertise as the globe becomes more and more competit

...more

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10 months ago

Sambalpur University Business Management Studies

What is the CPET date for admission to MBA offered by Sambalpur University?

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ABHINAV SRIVASTAVA

Contributor-Level 9

The Common PG Entrance Test (CPET) for MBA admission at Sambalpur University will be held on dates such as June 23, 24, 26, 27, 29, July 1, 3, and 4 in 2025. The application will begin in May, and the last date to submit the application form is around mid-May. The admit cards will be published to download in early June. The merit list is expected to be published in August, and the admission process will be done in September. The first-year MBA classes will start in September. Please verify official notifications!

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10 months ago

43. Find the vector equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane

$\vec{r} = (\hat{i} + 2 \hat{j} - 5 \hat{k}) + 9 = 0$

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Contributor-Level 10

The position vector of the point $(1, 2, 3)$ is $\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The direction ratios of the normal to the plane, $\vec{r} = (\hat{i} + 2 \hat{j} - 5 \hat{k}) + 9 = 0$ , are $1, 2, a n d - 5$ and the normal vector is $\vec{N} = (\hat{i} + 2 \hat{j} - 5 \hat{k})$

The equation of a line passing through a point and perpendicular to the given plane is given by,

$\vec{l} = \vec{r} + λ \vec{N}, λ \in R \Rightarrow \vec{l} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + 2 \hat{j} - 5 \hat{k})$

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10 months ago

Balaji Institute of Technology Business Management Studies

29. sin²6x – sin²4x = sin 2x sin 10x

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Payal Gupta

Contributor-Level 10

29. L.H.S $L.H.S = s i n^{2} 6 x - s i n^{2} 4 x .$

$= (s i n 6 x + s i n 4 x) (s i n 6 x - s i n 4 x) . [a^{2} - b^{2} = (a - b) (a + b)]$

So, L.H.S $L.H.S. = (2 s i n (\frac{6 x + 4 x}{2}) c o s (\frac{6 x - 4 x}{2})) (2 c o s (\frac{6 x + 4 x}{2}) s i n (\frac{6 x - 4 x}{2}))$

$= (2 s i n \frac{10 x}{2} c o s \frac{2 x}{2}) (2 c o s \frac{10 x}{2} s i n \frac{2 x}{2})$

$= 2 s i n 5 x c o s x \cdot 2 c o s 5 x s i n x$

$= 2 s i n 5 x c o s 5 x \cdot 2 c s i n x c o s x .$

As $s i n 2 θ = 2 s i n θ c o s θ$ we can write.

$L.H.S = s i n (2 \times 5 x) = s i n (2 \times x)$

$= s i n 10 x s i n 2 x = R.H.S.$ R.H.S

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10 months ago

What are the career options available after completing MBA course from Balaji Institute of Technology Management?

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Sanjana Dixit

Contributor-Level 10

MBA course is a two-year postgraduate level degree program focusing on Management, Business, and Entrepreneurial concepts. Students can check the list of career options available after completing MBA from Balaji Institute of Technology Management:

Job Profiles	Description	Average Salary (in INR)
Business Analyst	A Business Analyst is a professional who works within an organisation to analyze, understand, and document the business processes and systems.	9.5 LPA
Investment Banker	An Investment Banker is a finance professional who works within an investment bank or a financial institution, specializing in providing financial advisory services and raising capital for corporations, governments, and other entities.	5 LPA
Chief Financial Officer	The CFO is a high-level executive responsible for managing an organization's financial actions, strategies, and operations.	47.1 LPA

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10 months ago

42. If the lines $\frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ are perpendicular, find the value of k

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Contributor-Level 10

The direction of ratios of the lines, $\frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ , are $- 3, 2 k, 2 a n d 3 k, 1, - 5$ respectively.

It is known that two lines with direction ratios, $a_{1}, b_{1}, c_{1} a n d a_{2}, b_{2}, c_{2}$ , are perpendicular, if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$\begin{array}{l} ∴ - 3 (3 k) + 2 k \times 1 + 2 (- 5) = 0 \\ \Rightarrow - 9 k + 2 k - 10 = 0 \\ \Rightarrow 7 k = - 10 \\ \Rightarrow k = \frac{- 10}{7} \end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

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10 months ago

41. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

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Contributor-Level 10

The coordinates of $A, B, C, a n d D a r e (1, 2, 3), (4, 5, 7), ( - 4, 3, - 6),$ and $(2, 9, 2)$ respectively.

The direction ratios of $A B a r e (4 - 1) = 3, (5 - 2) = 3, a n d (7 - 3) = 4$

The direction ratios of $C D a r e (2 - (- 4)) = 6, (9 - 3) = 6, a n d (2 - (- 6)) = 8$

It can be seen that, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = \frac{1}{2}$

Therefore, AB is parallel to CD.

Thus, the angle between $A B a n d C D i s e i t h e r 0 ° o r 18$ $0$ $° .$

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10 months ago

40. Find the equation of the line parallel to x-axis and passing through the origin.

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Contributor-Level 10

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by $(a, 0, 0), w h e r e a ? R .$ Direction ratios of $O A a r e (a ? 0) = a, 0, 0$

The equation of OA is given by,

$\begin{array}{l} \frac{x ? 0}{a} = \frac{y ? 0}{0} = \frac{z ? 0}{0} \\ ? \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a \end{array}$

Thus, the equation of line parallel to x-axis and passing through origin is

$\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

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10 months ago

39. If $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1} .$

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Contributor-Level 10

It is given that $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines. Therefore,

$\begin{array}{l} l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 . . . . . . . . . . (1) \\ {l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2} = 1 . . . . . . . . . . (2) \\ {l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2} = 1 . . . . . . . . . . (3) \end{array}$

Let $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2} .$

$\begin{array}{l} ∴ l l_{1} + m m_{1} + n n_{1} = 0 \\ l l_{2} + m m_{2} + n n_{2} = 0 \\ ∴ \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - n_{2} l_{1}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{2})}^{2}} \\ = \frac{l^{2} + m^{2} + n^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{2})}^{2}} . . . . . . . . . . (4) \end{array}$

$l, m, n$ are the direction cosines of the line.

$∴ l^{2} + m^{2} + n^{2} = 1 \dots (5)$

It is known that,

$\begin{array}{l} ({l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2}) ({l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2}) - {(l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})}^{2} \\ = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ F r o m, (1), (2) & (3), w e . o b t a i n \\ \Rightarrow 1.1 - 0 = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ ∴ {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} = 1 . . . . . . . . . . (6) \end{array}$

Substituting the values from equations (5) and (6) in equation (4), we obtain

$\frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} = 1$

Thus, the direction cosines of the required line are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1}$

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10 months ago

$28.$ $c o s (\frac{3 π}{4} + x) - c o s (\frac{3 π}{4} - x) = -\sqrt2 s i n x$

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Payal Gupta

Contributor-Level 10

$28.$ L.H.S = $L.H.S = c o s (\frac{3 π}{4} + x) - c o s (\frac{3 π}{4} - x)$

Using cos (A + B) = cos A cos B – sin A sin B

and cos (A – B) = cos A cos B + sin A sin B

$L.H.S = [c o s \frac{3 π}{4} c o s x - s i n \frac{3 π}{4} s i n x] - [c o s \frac{3 π}{4} c o s x + s i n \frac{3 π}{4} s i n x]$

$= c o s \frac{3 π}{4} c o s x - s i n \frac{3 π}{4} s i n x - c o s \frac{3 π}{4} c o s x - s i n \frac{3 π}{4} s i n x .$

$= - 2 s i n \frac{3 π}{4} s i n x .$

$= - 2 s i n (\frac{4 π - π}{4}) s i n x .$

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10 months ago

38. Show that the line joining the origin to the point $(2, 1, 1)$ is perpendicular to the line determined by the points $(3, 5, - 1), (4, 3, - 1) .$

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Sambalpur University Colleges in India

Contributor-Level 10

Let OA be the line joining the origin, $O (0, 0, 0),$ and the point, $A (2, 1, 1) .$

Also, let BC be the line joining the points, $B (3, 5, - 1) a n d C (4, 3, - 1) .$

The direction ratios of $O A a r e 2, 1, a n d 1 a n d o f B C a r e (4 - 3) = 1, (3 - 5) = - 2, a n d (- 1 + 1) = 0$

OA is perpendicular to $B C, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$∴ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 1 + 1 (- 2) + 1 \times 0 = 2 - 2 = 0$

Thus, OA is perpendicular to BC.

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10 months ago

Who is eligible to apply for Sambalpur University MBA course?

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Shailja Rawat

Contributor-Level 10

Sambalpur University's MBA is a full-time management programme. For admissions to this programme, the university has a specified set of eligibility criteria. As per the criteria, students who have obtained a three-year bachelor's degree or a honours degree in any stream with a minimum of 50% aggregate are eligible to apply for the programme.

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10 months ago

Interior Design M.Sc

Which MSc Interior Design college offers the best placement options in India?

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Rashmi Srivastava

Contributor-Level 9

Here are some top MSc Interior Design college options offering great placement opportunities in India:

MSc Interior Design Colleges	Average Package Offered
Annamalai University Placements	INR 5.13 LPA
St. Joseph's College for Women Placements	INR 3.6 LPA
JEREC University Placements	INR 3.52 LPA
SNDT College Placements	INR 3.3 LPA

Note: The information is sourced from the official/counselling body of the institute/exam.

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10 months ago

Sambalpur University Business Management Studies

What is the total tuition fee for MBA course at Sambalpur University?

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Shailja Rawat

Contributor-Level 10

The total tuition fees for Sambalpur University MBA course ranges from INR 16,800 to INR 90,000. Apart from this, the total fees include other fee components, such as hostel fees, admission fees, etc.

Note: The above-mentioned fee is as per official sources. However, it is subject to change.

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10 months ago

St. Anne's Degree College for Women Surana College

How many courses are offered by St Annes Degree College for Women and Surana College for BBA?

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Aayush Sharma

Contributor-Level 10

Surana College and St Annes Degree College for Women both offer two BBA courses. The total tuition fee for the BBA programme at St Annes Degree College is INR 1.4 lakh, whereas for Surana College it is in the range of INR 3 Lacs - 3.6 Lakh. St Annes Degree College offers a BBA in Human Resource Management and BBA in Marketing and Finance. While, Surana College offers a BBA and BBA Professional course.

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10 months ago