Question

Prove that 1 + tan2a= sec2a

Answer 1

$</p><p>\begin{gathered}1 + {tan}^{2} = {sec}^{2} \\ = 1 + \frac{ {sin}^{2} }{ {cos}^{2} }(tan = \frac{sin}{cos} ) \\ = \frac{ {cos }^{2}+ {sin}^{2} }{cos {}^{2} }lcm \\ = \frac{1}{cos {}^{2} } {sin}^{2} + {cos}^{2} = 1 \\ = sec {}^{2} \\ hence \: proved\end{gathered}1+tan2=sec2=1+cos2sin2(tan=cossin)=cos2cos2+sin2lcm=cos21sin2+cos2=1=sec2 \\ \\ </p><p>$

Answer 2

Correct Question :

$\implies\sf 1 + Tan^{2}A = Sec^{2}A$

Solution :

Now,

We have to know that,

• $\implies\sf Tan A = \dfrac{Sin A}{Cos A}$

• $\implies\sf Sec A = \dfrac{1}{Cos A}$

• $\implies\sf Sin^{2} A + Cos^{2} A = 1$

Now, From above we know that,

$\implies\sf Tan A = \dfrac{Sin A}{Cos A}$

So,

$\implies\sf Tan^{2}A = \dfrac{Sin^{2} A}{Cos^{2} A}$

Now,

$\implies\sf 1 + \dfrac{Sin^{2} A}{Cos^{2} A}$

Taking L.C.M

$\implies\sf \dfrac{Cos^{2} A + Sin^{2} A}{Cos^{2} A}$

Using Trigonometry Identity :

$\implies\sf \dfrac{1}{Cos^{2} A}$

R.H.S side now,

$\dashrightarrow\:\:\sf Sec \: A = \dfrac{1}{Cos A}$

Then,

$\dashrightarrow\:\:\sf Sec^{2} \: A = \dfrac{1}{Cos^{2} A}$

Therefore,

$\dashrightarrow\:\: { \underline{ \boxed{\sf \dfrac{1}{Cos^{2} A} = \dfrac{1}{Cos^{2} A} }}}$

${\boxed{\bf{L.H.S = R.H.S}}}$

Hence, Proved !!