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Contributor-Level 10

The private Cyber Security colleges in Kerala, where students can study, are listed below along with their average course fees:

College Name	Average Fees
Amrita School of Engineering Admission	INR 3.18 Lakh- 24 Lakh
TOMS College of Engineering and Polytechnic Admission	INR 67,500
Boston Institute of Analytics Admission	INR 75,000- 1.55 Lakh
Rajadhani Institute of Engineering and Technology Admission	INR 2 Lakh
Sree Narayana Gurukulam College Of Engineering Admission	INR 2 Lakh

Disclaimer: This information is sourced from the official website and may vary.

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9 months ago

Which entrance exams can I take for admission to Kerala colleges that provide Cyber Security?

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Contributor-Level 10

The table below mentions some details about the dates, schedule, and syllabus of the entrance exams that are accepted at the GATE Security colleges in Kerala.

Exam Name	Exam Date	Exam Schedule	Exam Syllabus
KEAM	April, 2026 (Tentative)	KEAM Schedule	KEAM Syllabus
GATE	February, 2026 (Tentative)	GATE Schedule	GATE Syllabus
JEE MAIN	January, 2026 (Tentative)	JEE MAIN Schedule	JEE MAIN Syllabus
UGC NET	December 2025 (Tentative)	UGC NET Schedule	UGC NET Syllabus

Disclaimer: This information is sourced from the official website and may vary.

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9 months ago

Can I get some details about the Cyber Security colleges in Kerala?

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Contributor-Level 10

Here is some important information about the Cyber Security colleges in Kerala:

Particulars	Details
Number of Cyber Security Colleges in Kerala	30+ colleges
Annual Fees	Less than INR 1 Lakh: 5 colleges INR 1-2 Lakh: 5 colleges INR 2-3 Lakh: 5 colleges INR 3-5 Lakh: 3 colleges More than INR 5 Lakh: 3 colleges
Top Cyber Security Colleges in Kerala	Rajagiri College of Social Sciences, Indian Institute of Information Technology, Amrita School of Engineering, TOMS College of Engineering and Polytechnic, Boston Institute of Analytics, etc.
Accepted Entrance Exams	KEAM, GATE, JEE MAIN, UGC NET, etc.

Disclaimer: This information is sourced from the official website and may vary.

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9 months ago

Which are the top Cyber Security colleges in Kerala?

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Contributor-Level 10

Some of the top Cyber Security colleges in Kerala are Rajagiri College of Social Sciences, Indian Institute of Information Technology, Amrita School of Engineering, TOMS College of Engineering and Polytechnic, Boston Institute of Analytics, etc.

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9 months ago

How many Cyber Security colleges are there in Kerala?

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

Students can study at approximately 30+ Cyber Security colleges in Kerala, 23 private, 1 public-private, and 3 government, where students can take admission. Students are selected for these top Cyber Security colleges in Kerala based on the scores in entrance exams like KEAM, GATE, JEE MAIN, UGC NET, etc.

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9 months ago

If $t a n (π c o s θ) = c o t (π s i n θ)$ , then $c o s (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt[]{2}}$ .

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n (π c o s θ) = c o t (π s i n θ) \\ \Rightarrow t a n (π c o s θ) = t a n (\frac{π}{2} - π s i n θ) \\ \Rightarrow π c o s θ = \frac{π}{2} - π s i n θ \\ \Rightarrow π c o s θ + π s i n θ = \frac{π}{2} \\ \Rightarrow c o s θ + s i n θ = \frac{1}{2} \\ \Rightarrow \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ = \frac{1}{2 \sqrt[]{2}} \\ \Rightarrow c o s \frac{π}{4} c o s θ + s i n \frac{π}{4} s i n θ = \frac{1}{2 \sqrt[]{2}} \\ \Rightarrow c o s (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt[]{2}} [? c o s (θ - \frac{π}{4}) o r c o s (\frac{π}{4} - θ)] \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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9 months ago

If $z$ and $w$ are two complex numbers such that $? z w ? = 1$ and $arg (z) - arg (w) = \frac{π}{2}$ , then show that $z^{} w = - i$ .

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Vogue Institute of Art and Design Colleges in Bangalore

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d w = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ z w = r_{1} r_{2} [(c o s θ_{1} - i s i n θ_{1})] [(c o s θ_{2} + i s i n θ_{2})] \\ | z w | = r_{1} r_{2} = 1 [G i v e n] \\ N o w a r g (z) - a r g (w) = \frac{π}{2} \\ \Rightarrow θ_{1} - θ_{2} = \frac{π}{2} \Rightarrow a r g (\frac{z}{w}) = \frac{π}{2} \\ \bar{z} w = r_{1} (c o s θ_{1} - i s i n θ_{1}) r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} r_{2} [c o s θ_{1} c o s θ_{2} + i c o s θ_{1} s i n θ_{2} - i s i n θ_{1} c o s θ_{2} - i^{2} s i n θ_{1} s i n θ_{2}] \\ = r_{1} r_{2} [(c o s θ_{1} c o s θ_{2} + s i n θ_{1} s i n θ_{2}) + i (c o s θ_{1} s i n θ_{2} - s i n θ_{1} c o s θ_{2})] \\ = r_{1} r_{2} [c o s (θ_{2} - θ_{1}) + i s i n (θ_{2} - θ_{1})] \\ = r_{1} r_{2} [c o s (- \frac{π}{2}) + i s i n (- \frac{π}{2})] \\ = r_{1} r_{2} [c o s \frac{π}{2} - i s i n \frac{π}{2}] = 1 . [0 - i] = - i \\ H e r e \bar{z} w = - i . H e n c e p r o v e d . \end{array}$

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9 months ago

What are the various programs offered by Vogue Institute of Art and Design?

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Ashwin Yadav

Contributor-Level 10

Below are the various programs offered by Vogue Institute of Art and Design:

Bachelors Degree in Fashion & Apparel Designing – B.Sc (FAD)
Bachelors Degree in Interior Design & Decoration – B.Sc (IDD)
Bachelors Degree in Jewellery Design & Management – B.Sc (JDM)
Bachelors Degree in Visual Arts – BVA (A&GA)
Bachelor of Design in Fashion Design – B.Des. (FD)
Bachelor of Design in Interior Design – B. Des. (ID)
M.Des – Fashion Design Management (FDM) – 2 Years
M.Des – Interior Design Management (IDM) – 2 Years

Apart from these, the students are also offered various Diploma programs in the same disciplines.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $t a n θ + t a n 2 θ + 3 t a n θ t a n 2 θ = 3$ , then $θ = \frac{n π}{3} + \frac{π}{9}$ .

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ + t a n 2 θ + \sqrt[]{3} t a n θ t a n 2 θ = \sqrt[]{3} \\ \Rightarrow t a n θ + t a n 2 θ = \sqrt[]{3} - \sqrt[]{3} t a n θ t a n 2 θ \\ \Rightarrow t a n θ + t a n 2 θ = \sqrt[]{3} (1 - t a n θ t a n 2 θ) \\ \Rightarrow \frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} = \sqrt[]{3} \\ \Rightarrow t a n (θ + 2 θ) = \sqrt[]{3} \\ \Rightarrow t a n 3 θ = t a n \frac{π}{3} ∴ 3 θ = n π + \frac{π}{3} \\ S o, θ = \frac{n π}{3} + \frac{π}{9} \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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9 months ago

Write the complex number $z = \frac{1 - i}{c o s \frac{π}{3} + i s i n \frac{π}{3}}$ in polar form.

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \\ z = \frac{1 - i}{\frac{1}{2} + i \frac{\sqrt[]{3}}{2}} = \frac{2 - 2 i}{1 + i \sqrt[]{3}} = \frac{2 - 2 i}{1 + i \sqrt[]{3}} \times \frac{1 - i \sqrt[]{3}}{1 - i \sqrt[]{3}} \\ \Rightarrow z = \frac{2 - 2 \sqrt[]{3} i - 2 i + 2 \sqrt[]{3} i^{2}}{{(1)}^{2} - {(i \sqrt[]{3})}^{2}} = \frac{2 - 2 \sqrt[]{3} i - 2 i - 2 \sqrt[]{3}}{1 - 3 i^{2}} \\ = \frac{(2 - 2 \sqrt[]{3}) - (2 + 2 \sqrt[]{3}) i}{4} = \frac{1 - \sqrt[]{3}}{2} - \frac{1 + \sqrt[]{3}}{2} i \\ r = \sqrt[]{{(\frac{1 - \sqrt[]{3}}{2})}^{2} + {(- \frac{1 + \sqrt[]{3}}{2})}^{2}} = \sqrt[]{\frac{1 + 3 - 2 \sqrt[]{3}}{4} + \frac{1 + 3 + 2 \sqrt[]{3}}{4}} \\ = \sqrt[]{\frac{4 - 2 \sqrt[]{3} + 4 + 2 \sqrt[]{3}}{4}} = \sqrt[]{\frac{8}{4}} = \sqrt[]{2} \\ S o r = \sqrt[]{2} \\ N o w a r g (z) = t a n^{- 1} \frac{y}{x} \\ \Rightarrow θ = t a n^{- 1} \frac{- (\frac{1 + \sqrt[]{3}}{2})}{(\frac{1 - \sqrt[]{3}}{2})} = t a n^{- 1} [- (\frac{1 + \sqrt[]{3}}{1 - \sqrt[]{3}})] = t a n^{- 1} \frac{\sqrt[]{3} + 1}{\sqrt[]{3} - 1} \\ \Rightarrow θ = t a n^{- 1} [t a n (\frac{π}{4} + \frac{π}{6})] [? t a n (\frac{π}{4} + \frac{π}{6}) = \frac{t a n \frac{π}{4} + t a n \frac{π}{6}}{1 - t a n \frac{π}{4} t a n \frac{π}{6}}] \\ \Rightarrow θ = \frac{5 π}{12} \\ H e n c e, t h e p o l a r i s z = \sqrt[]{2} [c o s (\frac{5 π}{12}) + i s i n (\frac{5 π}{12})] . \end{array}$

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9 months ago

If $c o s e c x = 1 + c o t x$ , then $x = 2 n π, 2 n π + \frac{π}{2}$ .

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s e c x = 1 + c o t x \\ \Rightarrow \frac{1}{s i n x} = 1 + \frac{c o s x}{s i n x} \\ \Rightarrow \frac{1}{s i n x} = \frac{s i n x + c o s x}{s i n x} \\ \Rightarrow s i n x + c o s x = 1 \\ \Rightarrow \frac{1}{\sqrt[]{2}} s i n x + \frac{1}{\sqrt[]{2}} c o s x = \frac{1}{\sqrt[]{2}} \\ \Rightarrow s i n \frac{π}{4} s i n x + c o s \frac{π}{4} c o s x = \frac{1}{\sqrt[]{2}} \\ \Rightarrow c o s (x - \frac{π}{4}) = \frac{1}{\sqrt[]{2}} \\ \Rightarrow c o s (x - \frac{π}{4}) = c o s \frac{π}{4} \\ \Rightarrow x = 2 n π + \frac{π}{4} + \frac{π}{4} \Rightarrow x = 2 n π + \frac{π}{2} \\ o r x = 2 n π + \frac{π}{4} - \frac{π}{4} \Rightarrow x = 2 n π \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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9 months ago

Find the complex number satisfying the equation $z + \sqrt 2 ? z + 1 ? + i = 0$ .

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : z + \sqrt[]{2} | (z + 1) | + i = 0 \\ L e t z = x + y i \\ ∴ (x + y i) + \sqrt[]{2} | (x + y i + 1) | + i = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} | (x + 1) + y i | = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} \sqrt[]{{(x + 1)}^{2} + y^{2}} = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} = 0 + 0 i \\ \Rightarrow x + \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} = 0, y + 1 = 0 \\ \Rightarrow x = - \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} a n d y = - 1 \\ \Rightarrow x^{2} = 2 (x^{2} + 2 x + 1 + y^{2}) \\ \Rightarrow x^{2} = 2 x^{2} + 4 x + 2 + 2 y^{2} \\ \Rightarrow x^{2} + 4 x + 2 + 2 y^{2} = 0 \\ \Rightarrow x^{2} + 4 x + 2 + 2 {(- 1)}^{2} = 0 [? y = - 1] \\ \Rightarrow x^{2} + 4 x + 4 = 0 \\ \Rightarrow {(x + 2)}^{2} = 0 \\ \Rightarrow x + 2 = 0 \Rightarrow x = - 2 \\ H e n c e, z = x + y i = - 2 - i . \end{array}$

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9 months ago

One value of $θ$ which satisfies the equation $s i n^{4} θ - 2 s i n^{2} θ - 1 = 0$ lies between $0$ and $2 π$ .

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How would I know which MSc specialisation is the best for me after BSc Nursing?

Contributor-Level 10

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \\ s i n^{4} θ - 2 s i n^{2} θ - 1 = 0 \\ s i n^{2} θ = \frac{- (- 2) \pm \sqrt[]{{(- 2)}^{2} - 4 \times 1 \times - 1}}{2 \times 1} \\ = \frac{2 \pm \sqrt[]{4 + 4}}{2} = \frac{2 \pm \sqrt[]{8}}{2} = \frac{2 \pm 2 \sqrt[]{2}}{2} = 1 \pm \sqrt[]{2} \\ ∴ s i n^{2} θ = (1 + \sqrt[]{2}) o r (1 - \sqrt[]{2}) \Rightarrow - 1 \leq s i n θ \leq 1 \\ \Rightarrow s i n^{2} θ \leq 1 b u t s i n^{2} θ = (1 + \sqrt[]{2}) o r (1 - \sqrt[]{2}) \\ W h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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9 months ago

M.Sc. BSc Nursing Scope

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Shikha Arora

Contributor-Level 9

At the BSc Nursing level, you'll learn the basics of every specialization. Once you complete the degree, you can pursue an MSc Nursing in whichever specialization you want. By that time, you'll know the most suitable specialization for yourself, which'll excite you, and motivate you to pursue a career in.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Solve the system of equations $Re (z^{2}) = 0, z = 2$ .

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I would like to know about the online P. B. BSc Nursing course offered at Bengaluru City University. Can you help me with the same?

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : R e (z^{2}) = 0 a n d | z | = 2 \\ L e t z = x + y i \\ ∴ | z | = \sqrt[]{x^{2} + y^{2}} \\ \Rightarrow \sqrt[]{x^{2} + y^{2}} = 2 \Rightarrow x^{2} + y^{2} = 4 \dots (i) \\ Since z = x + y i \\ z^{2} = x^{2} + y^{2} i^{2} + 2 x y i \Rightarrow z^{2} = x^{2} - y^{2} + 2 x y i \\ ∴ R e (z^{2}) = x^{2} - y^{2} \\ \Rightarrow x^{2} - y^{2} = 0 \dots (i i) \\ F r o m e q n . (i) a n d (i i), w e g e t \\ x^{2} + y^{2} = 4 \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt[]{2} a n d y = \pm \sqrt[]{2} \\ H e n c e, z = \pm \sqrt[]{2} \pm i \sqrt[]{2}, - \sqrt[]{2} \pm i \sqrt[]{2} . \end{array}$

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9 months ago

online BSc Nursing

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

. $c o s \frac{π}{15} \cdot c o s \frac{4 π}{15} \cdot c o s \frac{8 π}{15} \cdot c o s \frac{16 π}{15} = \frac{1}{16}$

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Vogue Institute of Art and Design Colleges in Bangalore

Contributor-Level 10

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . c o s \frac{2 π}{15} . c o s \frac{4 π}{15} . c o s \frac{8 π}{15} . c o s \frac{16 π}{15} \\ = c o s 2 4^{0} . c o s 4 8^{0} . c o s 9 6^{0} . c o s 19 2^{0} \\ = \frac{1}{16 s i n 2 4^{0}} [(2 s i n 2 4^{0} c o s 2 4^{0}) (2 c o s 4 8^{0}) (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 4 8^{0} . 2 c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 4 8^{0} c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 9 6^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 9 6^{0} c o s 9 6^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 19 2^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 19 2^{0} c o s 19 2^{0}] \\ = \frac{1}{16 s i n 2 4^{0}} s i n 38 4^{0} = \frac{1}{16 s i n 2 4^{0}} s i n (36 0^{0} + 2 4^{0}) \\ = \frac{1}{16 s i n 2 4^{0}} \times s i n 2 4^{0} [? s i n (36 0^{0} + θ) = s i n θ] \\ = \frac{1}{16} R . H . S . \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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9 months ago

What is the eligibility criteria for BDes at Vogue Institute of Bangalore?

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Ashwin Yadav

Contributor-Level 10

Students seeking admissions to the BDes course at Vogue Institute of Art and Design must clear their Higher Secondary exams with the minimum 50% scoring. Students must also appear for the in-house Vogue Institute of Art and Design Aptitude Test. Candidates can visit the official website and learn more regarding the coruse-wise details.

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9 months ago

KAU - Kerala Agricultural University B.Sc.

What is the deadline to apply for BSc courses at KAU?

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Indrani Tyagi

Contributor-Level 10

The deadlines/ last date to apply for BSc courses at KAU are released for the academic batch of 2025. Additionally, the deadlines to apply for BSc courses and integrated BSc + MSc courses are different. For example, the deadlines to apply for BSc courses at KAU are usually released in the month of June, and for integrated courses, the deadlines are usually announced in the month of May.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

$s i n 1 0^{\circ}$ is greater than $c o s 1 0^{\circ}$ .

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