Prove that: tan theta - 1 …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Shalet Menezes 7 years, 3 months ago
- 1 answers
Test
Related Questions
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 1 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Hari Anand 6 months, 1 week 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
Sia ? 6 years, 6 months ago
L.H.S={tex}\frac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 }{/tex}
{tex}= \frac { ( \tan \theta + \sec \theta ) - \left( \sec ^ { 2 } \theta - \tan ^ { 2 } \theta \right) } { \tan \theta - \sec \theta + 1 }{/tex} {tex}[\because sec^2\theta-tan^2\theta=1]{/tex}
{tex}= \frac { ( \tan \theta + \sec \theta ) - ( \sec \theta - \tan \theta ) ( \sec \theta + \tan \theta ) } { \tan \theta - \sec \theta + 1 }{/tex}
{tex}= \frac { ( \tan \theta + \sec \theta ) [ 1 - \sec \theta + \tan \theta ] } { \tan \theta - \sec \theta + 1 }{/tex}
= {tex}tan\theta+sec\theta{/tex}
= R.H.S.
Hence Proved.
0Thank You