Ask & Answer: India's Largest Education Community

The placement salary of theNizam top BCA colleges in Hyderabad is given in the table below:

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

Yes! There are 3 government colleges in Hyderabad that offer the BCA course. Check the table provided below for details about these colleges:

Disclaimer: This College of Physical Education 

—

Merit-Based

Vivekananda Government Degree College Admission

—

Merit-Based

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

Read More: Best BCA Colleges in Hyderabad

Yes! There are 35+ private colleges in Hyderabad that offer the BCA course. Check the table provided below for details about these colleges:

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

Read More: Best BCA Colleges in Hyderabad 

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

$\begin{array}{l}Giventhat:\left(1+i\right)z=\left(1-i\right)\overline{z}\\ \Rightarrow \frac{z}{\overline{z}}=\frac{1-i}{1+i}=\frac{1-i}{1+i}\times \frac{1-i}{1-i}=\frac{1+i^2-2i}{1-i^2}\\ =\frac{1-1-2i}{1+1}=\frac{-2i}{2}=-i\\ \Rightarrow \frac{z}{\overline{z}}=-i\\ \therefore z=-i.\overline{z}Henceproved.\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:cos2\theta cos2?+si{n}^2\left(\theta -?\right)-si{n}^2\left(\theta +?\right)\\ cos2\theta cos2?+si{n}^2\left(\theta -?\right)-si{n}^2\left(\theta +?\right)=cos2\theta cos2?+sin\left(\theta -?+\theta +?\right).sin\left(\theta -?-\theta -?\right)\\ \left[?si{n}^{2}A-si{n}^{2}B=sin\left(A+B\right).sin\left(A-B\right)\right]\\ =cos2\theta cos2?+sin2\theta .sin\left(-2?\right)\\ =cos2\theta cos2?-sin2\theta .sin2?\left[?sin\left(-\theta \right)=-sin\theta \right]\\ =cos\left(2\theta +2?\right)=cos2\left(\theta +?\right)\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

The top entrance exams through which students can gain admission to the top BCA colleges of Hyderabad are mentioned in the table below:

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

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

$\begin{array}{l}Giventhat:a=cos\theta +isin\theta \\ \therefore \frac{1+a}{1-a}=\frac{1+cos\theta +isin\theta }{1-cos\theta -isin\theta }\\ =\frac{1+cos\theta +isin\theta }{1-cos\theta -isin\theta }\times \frac{1-cos\theta +isin\theta }{1-cos\theta +isin\theta }\\ =\frac{1-cos\theta +isin\theta +cos\theta -co{s}^2\theta +isin\theta cos\theta +isin\theta -isin\theta cos\theta +i^{2}si{n}^2\theta }{{\left(1-cos\theta \right)}^2-i^{2}si{n}^2\theta }\\ =\frac{1+isin\theta -co{s}^2\theta +isin\theta -si{n}^2\theta }{1+co{s}^2\theta -2cos\theta +si{n}^2\theta }\\ =\frac{si{n}^2\theta +2isin\theta -si{n}^2\theta }{1+1-2cos\theta }=\frac{2isin\theta }{2-2cos\theta }=\frac{2isin\theta }{2\left(1-cos\theta \right)}=\frac{isin\theta }{1-cos\theta }\\ =\frac{2sin\frac{\theta }{2}cos\frac{\theta }{2}.i}{2si{n}^2\frac{\theta }{2}}=cot\frac{\theta }{2}.i\\ Hence,\frac{1+a}{1-a}=i.cot\frac{\theta }{2}.\end{array}$

Yes, of course! Take a look at the table below to learn more about the top BCA colleges in Hyderabad.

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

There are many popular BCA colleges in Hyderabad where students can study. Some of the most well-known and well-reputed colleges include Nizam College, Woxsen University, Siva Sivani Degree College, Annie Besant Women's College, Rockwell Business School, London Management Academy, etc.

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenexpressioniscot\left(\frac{\pi }{4}+\theta \right).cot\left(\frac{\pi }{4}-\theta \right)\\ =\frac{cot\frac{\pi }{4}cot\theta -1}{cot\theta +cot\frac{\pi }{4}}\times \frac{cot\frac{\pi }{4}cot\theta +1}{cot\theta -cot\frac{\pi }{4}}\\ =\frac{1.cot\theta -1}{cot\theta +1}\times \frac{1.cot\theta +1}{cot\theta -1}\\ =\frac{cot\theta -1}{cot\theta +1}\times \frac{cot\theta +1}{cot\theta -1}=1\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

Hi,

There are around 50+ BCA colleges in Hyderabad where students can study. This includes 39 private, 3 government, and 3 public-private colleges. Students can take admission to these colleges based on their Class 12 marks or through their scores in the CUET entrance exam.

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

$\begin{array}{l}Wehave{\left(\frac{1-i}{1+i}\right)}^{100}=a+bi\\ \Rightarrow {\left(\frac{1-i}{1+i}\times \frac{1-i}{1-i}\right)}^{100}=a+bi\Rightarrow {\left(\frac{1+i^2-2i}{1-i^2}\right)}^{100}=a+bi\\ \Rightarrow {\left(\frac{1-1-2i}{1+1}\right)}^{100}=a+bi\Rightarrow {\left(\frac{-2i}{2}\right)}^{100}=a+bi\\ \Rightarrow {\left(-i\right)}^{100}=a+bi\Rightarrow i^{100}=a+bi\\ \Rightarrow {\left(i^4\right)}^{25}=a+bi\Rightarrow {\left(1\right)}^{25}=a+bi\Rightarrow 1=a+bi\\ \Rightarrow 1+0i=a+bi\\ Comparingtherealandimaginaryparts,wegeta=1,b=0.\\ Hence,\left(a,b\right)=\left(1,0\right).\end{array}$

The fee structure for the PK University BA programme is available on the university's official website. However, the mentioned fee is indicative and may be subject to change by the university or the sanctioning body. Applicants should regularly check the site for the most up-to-date information.

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

$\begin{array}{l}Wehave\frac{{\left(1+i\right)}^2}{2-i}=x+iy\\ \Rightarrow \frac{1+i^2+2i}{2-i}=x+iy\Rightarrow \frac{1-1+2i}{2-i}=x+iy\\ \Rightarrow \frac{2i}{2-i}=x+iy\Rightarrow \frac{2i}{2-i}\times \frac{2+i}{2+i}=x+iy\\ \Rightarrow \frac{4i+2i^2}{4-i^2}=x+iy\Rightarrow \frac{4i-2}{4+1}=x+iy\left[?i^2=-1\right]\\ \Rightarrow \frac{-2+4i}{5}=x+iy\Rightarrow \frac{-2}{5}+\frac{4}{5}i=x+iy\\ Comparingtherealandimaginaryparts,wegetx=\frac{-2}{5},y=\frac{4}{5}.\\ Hence,x+y=\frac{-2}{5}+\frac{4}{5}=\frac{2}{5}.\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenexpressionissin\left(45^0+\theta \right)-cos\left(45^0-\theta \right)\\ sin\left(45^0+\theta \right)=sin45^{0}cos\theta +cos45^{0}sin\theta \\ =\frac{1}{\sqrt[]{2}}cos\theta +\frac{1}{\sqrt[]{2}}sin\theta \\ cos\left(45^0-\theta \right)=cos45^{0}cos\theta +sin45^{0}sin\theta \\ =\frac{1}{\sqrt[]{2}}cos\theta +\frac{1}{\sqrt[]{2}}sin\theta \\ sin\left(45^0+\theta \right)-cos\left(45^0-\theta \right)=\frac{1}{\sqrt[]{2}}cos\theta +\frac{1}{\sqrt[]{2}}sin\theta -\frac{1}{\sqrt[]{2}}cos\theta -\frac{1}{\sqrt[]{2}}sin\theta =0\\ Hence,thecorrectoptionis\left(d\right).\end{array}$