If tan +sec=l then prove that …
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Sia ? 6 years, 6 months ago
Given, tan θ + sec θ = {tex}l{/tex}........(1)
We know that, sec2 θ – tan2 θ = 1.......(2)
Now, sec θ + tan θ = {tex}l{/tex} [ from (1) ]
⇒ (sec θ + tan θ) {tex}\frac { ( \sec \theta - \tan \theta ) } { \sec \theta - \tan \theta } = 1{/tex}
⇒ {tex}\frac { \sec ^ { 2 } \theta - \tan ^ { 2 } \theta } { \sec \theta - \tan \theta } = l{/tex}
⇒ {tex}\frac { 1 } { \sec \theta - \tan \theta } = l{/tex} [ from equation (2) ]
or, sec θ – tan θ = {tex}\frac { 1 } { l }{/tex} ........(3)
Now, to get sec θ , eliminating tan θ from (1) and (3)
adding (1) and (3) we get :-
⇒ 2 sec θ = {tex}l + \frac { 1 } { l }{/tex}
⇒ 2 sec θ = {tex}\frac { l ^ { 2 } + 1 } { l }{/tex}
⇒ sec θ = {tex}\frac { l ^ { 2 } + 1 } { 2 l }{/tex}
Hence, proved.
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