sec² thita-cos² = tan² thita + sin² thita
Share
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
[tex]\displaystyle{\boxed{\red{\sf\:\sec^2\:\theta\:-\:\cos^2\:\theta\:=\:\tan^2\:\theta\:+\:\sin^2\:\theta}}}[/tex]
Step-by-step-explanation:
We have given a trigonometric equation.
We have to prove that trigonometric equation.
The given trigonometric equation is
[tex]\displaystyle{\sf\:\sec^2\:\theta\:-\:\cos^2\:\theta\:=\:\tan^2\:\theta\:+\:\sin^2\:\theta}[/tex]
[tex]\displaystyle{\sf\:LHS\:=\:\sec^2\:\theta\:-\:\cos^2\:\theta}[/tex]
[tex]\displaystyle{\implies\sf\:LHS\:=\:\sec^2\:\theta\:-\:(\:1\:-\:\sin^2\:\theta\:)\:\quad\:-\:-\:-\:[\:\because\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1\:]}[/tex]
[tex]\displaystyle{\implies\sf\:LHS\:=\:\sec^2\:\theta\:-\:1\:+\:\sin^2\:\theta}[/tex]
[tex]\displaystyle{\implies\sf\:LHS\:=\:1\:+\:\tan^2\:\theta\:-\:1\:+\:\sin^2\:\theta\:\quad\:-\:-\:-\:[\:\because\:\sec^2\:\theta\:=\:1\:+\:\tan^2\:\theta\:]}[/tex]
[tex]\displaystyle{\implies\sf\:LHS\:=\:\tan^2\:\theta\:-\:\cancel{1}\:+\:\cancel{1}\:+\:\sin^2\:\theta}[/tex]
[tex]\displaystyle{\implies\sf\:LHS\:=\:\tan^2\:\theta\:+\:\sin^2\:\theta}[/tex]
[tex]\displaystyle{\implies\sf\:RHS\:=\:\tan^2\:\theta\:+\:\sin^2\:\theta}[/tex]
[tex]\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:LHS\:=\:RHS}}}}[/tex]
Hence proved!
Given : [tex]\sf{\sec^{2} \theta- \cos^{2} \theta = \tan^{2}\theta + \sin^{2}\theta }\\[/tex]
To Prove: [tex]\sf{\sec^{2} \theta- \cos^{2} \theta = \tan^{2}\theta + \sin^{2}\theta }\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀⠀⠀[tex]\underline {\sf{\star\:\sec^{2} \theta- \cos^{2} \theta = \tan^{2}\theta + \sin^{2}\theta }}\\[/tex]
Here :
⠀⠀⠀⠀⠀[tex]\implies \sf{L.H.S= \:\sec^{2} \theta- \cos^{2} \theta }\\\\[/tex]
⠀⠀⠀⠀⠀[tex]\implies \sf{R.H.S =\: \tan^{2}\theta + \sin^{2}\theta }\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\bf{\star\:Now \: By \: Solving \: the \: L.H.S \: \::}}\\[/tex]
⠀⠀⠀⠀⠀[tex]:\implies \tt{L.H.S= \:\sec^{2} \theta- \cos^{2} \theta }\\\\[/tex]
⠀⠀⠀⠀⠀[tex]\implies \sf{\:\sec^{2} \theta- \cos^{2} \theta }\\\\[/tex]
As , We know that :
⠀⠀⠀⠀⠀[tex]\implies \sf{\:\sec^{2} \theta-\purple {\cos^{2} \theta } }\\\\[/tex]
⠀⠀⠀⠀⠀[tex]\implies \sf{\:\sec^{2} \theta-\purple {1 + \sin^{2} \theta } }\\\\[/tex]
⠀⠀⠀⠀⠀[tex]\implies \sf{\:\sec^{2} \theta-1 + \sin^{2} \theta }\\\\[/tex]
As , We know that :
⠀⠀⠀⠀⠀[tex]\implies \sf{\:\purple {\sec^{2} \theta}-1 + \sin^{2} \theta }\\\\[/tex]
⠀⠀⠀⠀⠀[tex]\implies \sf{\:\purple {1 + \tan^{2} \theta}-1 + \sin^{2} \theta }\\\\[/tex]
⠀⠀⠀⠀⠀[tex]\implies \sf{\:\cancel {1} + \tan^{2} \theta \cancel{-1} + \sin^{2} \theta }\\\\[/tex]
⠀⠀⠀⠀⠀[tex]\underline {\boxed{\pink{ \mathrm { L.H.S\:= \: tan^{2} \theta + \sin^{2}\theta }}}}\:\bf{\bigstar}\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
Therefore,
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀[tex]\therefore \underline {\bf{ Hence \:Proved \:}}\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀