If sec theta +tan theta =p …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Prisha Singh 4 years, 10 months ago
- 1 answers
Related Questions
Posted by Anmol Kaushal 2 days, 3 hours ago
- 0 answers
Posted by Bhavan King 4 days, 13 hours ago
- 1 answers
Posted by Sahithi Bulusu 12 hours ago
- 0 answers
Posted by Jasmine Kaur 2 days, 20 hours ago
- 0 answers
Posted by Kangna Gautam 1 day, 23 hours ago
- 0 answers
Posted by Ritika Jain 1 day, 16 hours ago
- 0 answers
Posted by Udaya Peethala 2 days, 22 hours 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 ? 4 years, 10 months ago
{tex}sec\ \theta+ tan\ \theta = p{/tex} ...(i)
Also {tex}sec^2 \theta - tan^2 \theta = 1{/tex}
{tex}\Rightarrow{/tex} (sec {tex}\theta{/tex} - tan {tex}\theta{/tex}) (sec {tex}\theta{/tex} + tan {tex}\theta{/tex}) = 1
{tex}\Rightarrow{/tex} p(sec {tex}\theta{/tex} - tan {tex}\theta{/tex}) = 1
[using equation (i)]
{tex}\Rightarrow{/tex} sec {tex}\theta{/tex} - tan {tex}\theta{/tex} {tex}=\frac{1}{p}{/tex} ...(ii)
(ii) - (i) we get
{tex}-2 tan{/tex} {tex}\theta{/tex} {tex}=\frac{1-p^{2}}{p}{/tex}
{tex}\Rightarrow{/tex}- tan {tex}\theta{/tex} {tex}=\frac{1-p^{2}}{2 p}{/tex}
{tex}\Rightarrow{/tex}- cot {tex}\theta{/tex} {tex}=\frac{2 p}{1-p^{2}}{/tex}
cot {tex}\theta{/tex} {tex}=\left(\frac{2 p}{1-p^{2}}\right)^{2}{/tex}
{tex}=\frac{-4 p^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}\Rightarrow{/tex} {tex}cosec^2{/tex} {tex}\theta{/tex} - 1 {tex}=\frac{-4 p^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}\Rightarrow{/tex} {tex}cosec^2{/tex} {tex}\theta{/tex} {tex}=\frac{-4 p^{2}}{\left(1-p^{2}\right)^{2}}+1=\frac{-4 p^{2}+\left(1-p^{2}\right)^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}cosec^2{/tex} {tex}\theta{/tex} {tex}=\frac{-4 p^{2}+1+p^{4}-2 p^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}=\frac{\left(1+p^{2}\right)^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}\Rightarrow{/tex} {tex}cosec\ \theta{/tex} {tex}=\frac{1+p^{2}}{1-p^{2}}{/tex}
0Thank You