tan A1−cotA+cotA1−tanA Can be written as

sin2AcosA(sinA−cosA)+cos2AsinA(cosA−sinA)=sin3A−cos3AsinAcosA(sinA−cosA) sin2AcosA(sinA−cosA)+cos2AsinA(cosA−sinA)=1+sinAcosAsinAcosA sin2AcosA(sinA−cosA)+cos2AsinA(cosA−sinA)=secAcosecA+1

tan A1−cotA+cotA1−tanA Can be written as

sin2AcosA(sinA−cosA)+cos2AsinA(cosA−sinA)=sin3A−cos3AsinAcosA(sinA−cosA) sin2AcosA(sinA−cosA)+cos2AsinA(cosA−sinA)=1+sinAcosAsinAcosA sin2AcosA(sinA−cosA)+cos2AsinA(cosA−sinA)=secAcosecA+1

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ Can be written as

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\sec A \cos e c A + 1$

$\tan A + \cot A$

$\sec A + \cos e c A$

$\sin A \cos A + 1$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{\sin^{2} A}{\cos A (\sin A - \cos A)} + \frac{\cos^{2} A}{\sin A (\cos A - \sin A)} = \frac{\sin^{3} A - \cos^{3} A}{\sin A \cos A (\sin A - \cos A)}$
$\frac{\sin^{2} A}{\cos A (\sin A - \cos A)} + \frac{\cos^{2} A}{\sin A (\cos A - \sin A)} = \frac{1 + \sin A \cos A}{\sin A \cos A}$
$\frac{\sin^{2} A}{\cos A (\sin A - \cos A)} + \frac{\cos^{2} A}{\sin A (\cos A - \sin A)} = \sec A \cos e c A + 1$