[ 1-tanA]^2/[1-cotA]^2=tan^A
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Supriya Sahoo 6 years, 11 months ago
- 1 answers
Test
Related Questions
Posted by Kanika . 1 month, 1 week ago
- 1 answers
Posted by Lakshay Kumar 1 year, 1 month ago
- 0 answers
Posted by Hari Anand 6 months, 2 weeks ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 1 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Sahil Sahil 1 year, 4 months ago
- 2 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
Ram Kushwah 6 years, 11 months ago
{tex}\begin{array}{l}\frac{(1-\mathrm{tanA})^2}{(1-\mathrm{cotA})^2}=\frac{(1-{\displaystyle\frac{\mathrm{sinA}}{\mathrm{cosA}}})^2}{(1-\frac{\mathrm{cosA}}{\mathrm{sinA}})^2}\\=\frac{{\displaystyle(}\cos{\displaystyle\mathrm A}{\displaystyle-}{\displaystyle\mathrm{sinA}}{\displaystyle{\displaystyle)}^2}}{{\displaystyle\cos^2}{\displaystyle\mathrm A}}\times\frac{{\displaystyle\sin^2}{\displaystyle\mathrm A}}{(\mathrm{sinA}-\mathrm{cosA})^2}=\frac{\sin^2\mathrm A}{{\displaystyle\cos}^2\mathrm A}=\tan^2\mathrm A\\\end{array}{/tex}
4Thank You