Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
88 views
in Trigonometry by (239k points)
closed by
What is tan A + sec A equal to?
1. \(\rm \tan \left( \frac {\pi} 4 - \frac{A}{2} \right)\)
2. \(\rm \cot \left( \frac {\pi} 4 - \frac{A}{2} \right)\)
3. \(\rm 2\tan \left( \frac {\pi} 4 - \frac{A}{2} \right)\)
4. \(\rm 2\cot \left( \frac {\pi} 4 - \frac{A}{2} \right)\)

1 Answer

0 votes
by (237k points)
selected by
 
Best answer
Correct Answer - Option 2 : \(\rm \cot \left( \frac {\pi} 4 - \frac{A}{2} \right)\)

Concept:

  • sin 2x = 2 sin x cos x
  • cos 2x = cos2 x - sin2 x

 

Calculation:

Consider, tan A + sec A.

\(\rm \frac {\sin A}{\cos A} + \frac {1}{\cos A}\)

\(\rm \frac {\sin A+1}{\cos A}\)

\(\rm \frac {2\sin \frac{A}{2}\cos \frac{A}{2}+\sin^2\frac{A}{2}+\cos^2\frac{A}{2}}{\cos^2 \frac{A}{2}-\sin^2\frac{A}{2}}\)

\(\rm \frac {\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)^2}{\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)\left(\cos \frac{A}{2}-\sin \frac{A}{2}\right)}\)

\(\rm \frac {\sin \frac{A}{2}+\cos \frac{A}{2}}{\cos \frac{A}{2}-\sin \frac{A}{2}}\)

Dividing by \(\rm \sin\frac{A}{2}\), we get:

\(\rm \frac {\cot\frac{A}{2}+1}{\cot\frac{A}{2}-1}\)

Using \(\rm \cot\frac{\pi}{4}=1\), it can be written as;

\(\rm \frac {\cot\frac{\pi}{4}\cot\frac{A}{2}+1}{\cot\frac{A}{2}-\cot\frac{\pi}{4}}\)

\(\rm \cot \left( \frac {\pi} 4 - \frac{A}{2} \right)\)

No related questions found

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...