Question

Prove that : tan (45+A)= sec2A + tan2A

Answer 1

SOLUTION

TO PROVE

tan (45° + A) = sec2A + tan2A

PROOF

RHS

$= \sf{ \sec 2A\: + \tan 2A \: }$

$\displaystyle \sf{ = \frac{1}{ \cos 2A} + \frac{ \sin 2A}{ \cos 2A} \: }$

$\displaystyle \sf{ = \frac{1 + \sin 2A}{ \cos 2A} \: }$

$\displaystyle \sf{ = \frac{ { \sin}^{2} A+ { \cos}^{2} A+ 2\sin A \cos A}{ { \cos}^{2} A - { \sin}^{2} A} \: }$

$\displaystyle \sf{ = \frac{ { (\sin A + \cos A\: )}^{2} }{ ( { \cos} A + { \sin} A) \: ( { \cos} A - { \sin} A)}}$

$\displaystyle \sf{ = \frac{ { (\sin A + \cos A\: )} }{ ( { \cos} A - { \sin} A)}}$

Dividing numerator and denominator both by cosA

$\displaystyle \sf{ = \frac{ { \tan A +1\: } }{ 1 - { \tan} A}}$

$\displaystyle \sf{ = \frac{ { 1 + \tan A \: } }{ 1 - { \tan} A}}$

$\displaystyle \sf{ = \frac{ { \tan {45}^{ \circ} + \tan A \: } }{ 1 - \tan {45}^{ \circ} { \tan} A}}$

$\displaystyle \sf{ = { \tan( {45}^{ \circ} + A) \: } }$

= LHS

Hence proved

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If 3x=cosecθ and 3/x=cotθ,

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Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{\mathrm{Prove\:}\tan \left(45^{\circ \:}+a\right)=\sec \left(2a\right)+\tan \left(2a\right)}}$

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

$\bf{\tan \left(45^{\circ \:}+a\right)=\sec \left(2a\right)+\tan \left(2a\right)}$

We Have :

L.H.S = $\sf{\tan \left(45^{\circ \:}+a\right)}$

R.H.S = $\sf{\sec \left(2a\right)+\tan \left(2a\right)}$

$\mathrm{Manipulating\:left\:side}$

$\sf{\tan \left(45^{\circ \:}+a\right)}$

____________________

$\bf{\tan \left(45^{\circ \:}+a\right)}$

$\begin{aligned}&\sf{\text { Use the following identity: } \tan (x)=\frac{\sin (x)}{\cos (x)}}\\\\&\sf{=\frac{\sin \left(45^{\circ}+a\right)}{\cos \left(45^{\circ}+a\right)}}\\\\\\&\sf{\text { Use the following identity: } \sin (s+t)=\cos (s) \sin (t)+\cos (t) \sin (s)}\\\\&\sf{=\frac{\cos \left(45^{\circ}\right) \sin (a)+\cos (a) \sin \left(45^{\circ}\right)}{\cos \left(45^{\circ}+a\right)}}\end{aligned}$

$\begin{aligned}&\sF{\text { Use the Following identity: } \cos (s+t)=\cos (s) \cos (t)-\sin (s) \sin (t)}\\\\&\sf{=\frac{\cos \left(45^{\circ}\right) \sin (a)+\cos (a) \sin \left(45^{\circ}\right)}{\cos \left(45^{\circ}\right) \cos (a)-\sin \left(45^{\circ}\right) \sin (a)}}\end{aligned}$

$\displaystyle\sf{=\frac{\sin \left(a\right)+\cos \left(a\right)}{\cos \left(a\right)-\sin \left(a\right)}}$

$\boxed{\displaystyle\sf{\bf{\tan \left(45^{\circ \:}+a\right)}=\frac{\sin \left(a\right)+\cos \left(a\right)}{\cos \left(a\right)-\sin \left(a\right)}}}$

____________________

Now the expression becomes :-

Prove that :  $\displaystyle\sf{\frac{\sin \left(a\right)+\cos \left(a\right)}{\cos \left(a\right)-\sin \left(a\right)}=\sec \left(2a\right)+\tan \left(2a\right)}$

Now We Have :

L.H.S = $\sf{\dfrac{\sin \left(a\right)+\cos \left(a\right)}{\cos \left(a\right)-\sin \left(a\right)}}$

R.H.S = $\sf{\sec \left(2a\right)+\tan \left(2a\right)}$

____________________

$\mathrm{Manipulating\:right\:side}$

$\sf{\sec \left(2a\right)+\tan \left(2a\right)}$

$\displaystyle\sf{=\frac{1}{\cos \left(2a\right)}+\frac{\sin \left(2a\right)}{\cos \left(2a\right)}}$

$\sf{=\dfrac{1+\sin \left(2a\right)}{\cos \left(2a\right)}}$

$\mathrm{Use\:the\:following\:identity:}\:1+\sin \left(2x\right)=\left(\cos \left(x\right)+\sin \left(x\right)\right)^2$

$\mathrm{Use\:the\:following\:identity:}\:\cos \left(2x\right)=\cos ^2\left(x\right)-\sin ^2\left(x\right)$

$\displaystyle\sf{=\frac{\left(\cos \left(a\right)+\sin \left(a\right)\right)^2}{\cos ^2\left(a\right)-\sin ^2\left(a\right)}}$

$\displaystyle\sf{=\frac{\left(\cos \left(a\right)+\sin \left(a\right)\right)^2}{\left(\cos \left(a\right)+\sin \left(a\right)\right)\left(\cos \left(a\right)-\sin \left(a\right)\right)}}$

$\displaystyle\sf{=\frac{\cos \left(a\right)+\sin \left(a\right)}{\cos \left(a\right)-\sin \left(a\right)}}$

$\large\boxed{\displaystyle\sf{\frac{\sin \left(a\right)+\cos \left(a\right)}{\cos \left(a\right)-\sin \left(a\right)}\displaystyle\sf{=\frac{\cos \left(a\right)+\sin \left(a\right)}{\cos \left(a\right)-\sin \left(a\right)}}}}$

L.H.S = R.H.S

Hence Proved !!!