Question

$\sf \red{Prove \: that :- }$
$\sf{{ ( \frac{1 + \tan^{2} A }{1 + \cot^{2} A } )}= ( \frac{1 - \tan \: A }{1 - \cot A} ) ^{2} = \tan^{2} A}$
​

Answer 1

Question:-

$\\\\\sf {Prove \: that :- }$

$\\\sf{{ ( \frac{1 + \tan^{2} A }{1 + \cot^{2} A } )}= ( \frac{1 - \tan \: A }{1 - \cot A} ) ^{2} = \tan^{2} A}$

Required Answer:-

$\\\\ \sf{ \large{ \underline{ \underline{Formulas \: Applied}}}}$

$\\ \sf{ \bold{1 + \tan {}^{2} A} = \sec {}^{2} A}$

$\sf{ \bold{1 + \cot {}^{2} A }= \cosec {}^{2} A}$

$\sf{ \bold{ \sec\: A} = \frac{1}{ \cos \:A} }$

$\sf{ \bold{ \cosec \: A= \frac{1}{ \sin \:A } }}$

$\sf{ \bold{ \frac{ \sin {}^{2} A}{ \cos {}^{2} A } } = \tan {}^{2} A}$

$\sf{ \bold{ \cot \: A = \frac{1}{ \tan \: A } }}$

♡Solution :-

L. H. S. =

$\\ \sf{ \bold{ \frac{1 + \tan {}^{2}A }{1 + \cot {}^{2}A } }}$

$\\ \sf{ \implies{ \frac{ \sec {}^{2}A}{ \cosec{}^{2}A } }}$

$\\ \sf{ \large{ \implies{ \frac{ \frac{1}{ \cos {}^{2} A} }{ \frac{1}{ \sin {}^{2}A } } }}}$

$\\ \sf{ \implies{ \frac{ \sin {}^{2}A }{ \cos {}^{2} A } }}$

$\\ \sf{ \implies{ \pink{ \tan {}^{2} A}}}$

R. H. S. =

$\\ \sf{ \bold{( \frac{1 - \tan \: A }{1 - \cot \: A }) {}^{2} }}$

$\\ \sf{ \implies{ \large{ ( }\frac{1 - \tan \: A }{1 - \frac{1}{ \tan \: A} }} \large{) {}^{2} }}$

$\\ \sf{ \implies{ (\frac{1 - \tan \: A }{ \frac{ \tan \: A - 1 }{ \tan \: A}}) {}^{2} } }$

$\\ \sf{ \implies{( \frac{1 - \tan \: A }{ \tan \: A - 1 } \times \tan \: A}) {}^{2} }$

$\\ \sf{ \implies{ { (- \tan \: A) }^{2} }}$

$\\ \sf{ \implies{ \pink{ \tan {}^{2} \: A }}}$

$\\ \sf{ \green{ \bold{L.H.S.= R. H. S.=\tan {}^{2} \: A }}}$

Hence Proved!!

Answer 2

Answer:

sweet_quail

Step-by-step explanation:

ᴍᴀᴛʜᴇᴍᴀᴛɪᴄꜱ ɪꜱ ᴛʜᴇ ꜱᴄɪᴇɴᴄᴇ ᴛʜᴀᴛ ᴅᴇᴀʟꜱ ᴡɪᴛʜ ᴛʜᴇ ʟᴏɢɪᴄ ᴏꜰ ꜱʜᴀᴘᴇ, Qᴜᴀɴᴛɪᴛʏ ᴀɴᴅ ᴀʀʀᴀɴɢᴇᴍᴇɴᴛ. ᴍᴀᴛʜ ɪꜱ ᴀʟʟ ᴀʀᴏᴜɴᴅ ᴜꜱ, ɪɴ ᴇᴠᴇʀʏᴛʜɪɴɢ ᴡᴇ ᴅᴏ. ɪᴛ ɪꜱ ᴛʜᴇ ʙᴜɪʟᴅɪɴɢ ʙʟᴏᴄᴋ ꜰᴏʀ ᴇᴠᴇʀʏᴛʜɪɴɢ ɪɴ ᴏᴜʀ ᴅᴀɪʟʏ ʟɪᴠᴇꜱ, ɪɴᴄʟᴜᴅɪɴɢ ᴍᴏʙɪʟᴇ ᴅᴇᴠɪᴄᴇꜱ, ᴀʀᴄʜɪᴛᴇᴄᴛᴜʀᴇ (ᴀɴᴄɪᴇɴᴛ ᴀɴᴅ ᴍᴏᴅᴇʀɴ), ᴀʀᴛ, ᴍᴏɴᴇʏ, ᴇɴɢɪɴᴇᴇʀɪɴɢ, ᴀɴᴅ ᴇᴠᴇɴ ꜱᴘᴏʀᴛꜱ.