(1+cotA+tanA)(sinA-cosA)=secA/cosec2A-cosecA/sec2A
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Himanshu Goyal 6 years, 9 months ago
- 2 answers
Rashmi Bajpayee 6 years, 9 months ago
L.H.S.
{tex}\left( {1 + \cot {\rm{A}} + \tan {\rm{A}}} \right)\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right){/tex}
= {tex}\left( {1 + {{\cos {\rm{A}}} \over {\sin {\rm{A}}}} + {{\sin {\rm{A}}} \over {\cos {\rm{A}}}}} \right)\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right){/tex}
= {tex}{{\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right)\left( {\sin {\rm{AcosA}} + {{\sin }^2}{\rm{A}} + {{\cos }^2}{\rm{A}}} \right)} \over {\sin {\rm{AcosA}}}}{/tex}
= {tex}{{{{\sin }^3}{\rm{A}} - {{\cos }^3}{\rm{A}}} \over {\sin {\rm{AcosA}}}}{/tex}
= {tex}{{{{\sin }^2}{\rm{A}}} \over {{\rm{cosA}}}} - {{{{\cos }^2}{\rm{A}}} \over {\sin {\rm{A}}}}{/tex}
= {tex}{1 \over {{\rm{cosA}}}}.{{{{\sin }^2}{\rm{A}}} \over 1} - {1 \over {{\rm{sinA}}}}.{{{{\cos }^2}{\rm{A}}} \over 1}{/tex}
= {tex}{{\sec {\rm{A}}} \over {\cos e{c^2}{\rm{A}}}} - {{\cos ec{\rm{A}}} \over {{{\sec }^2}{\rm{A}}}}{/tex}
= R.H.S.
Related Questions
Posted by Ayush Alpha 1 month, 1 week ago
- 0 answers
Posted by Adi Manav 1 month, 1 week ago
- 0 answers
Posted by Ayush Alpha 1 month, 1 week ago
- 0 answers
Posted by Not Fine Bro 🎭 1 month ago
- 0 answers
Posted by Aditya Vinayak 1 month, 1 week ago
- 0 answers
Posted by Shraddha Mathur 4 weeks, 1 day ago
- 0 answers
Posted by Anant Gautam 1 week, 4 days ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Nidhi Pal 4 years, 1 month ago
2Thank You