Question

Prove that :
Sec^2 - cos^2 = sin^2 [sec^2 +1]

Answer 1

Correct question :-

Prove that :

sec² θ - cos² θ = sin² θ ( sec² θ + 1 )

Solution :-

$\sec^{2} \theta - \cos^{2} \theta = \sin^{2} \theta(sec^{2} \theta +1)$

$\tt{consider \ lhs} \\ \\ \sec^{2} \theta - \cos^{2} \theta \\ \\ \\ = \dfrac{1}{ \cos^{2} \theta } - \cos^{2} \theta \\ \\ \\ \{ \because \sec \theta = \dfrac{1}{ \cos \theta} \}$

$\tt{taking \ lcm} \\ \\ \\ = \dfrac{1 - \cos^{2} \theta( \cos ^{2} \theta)}{ \cos^{2} \theta} \\ \\ \\ = \dfrac{1^{2} - (\cos^{2} \theta)^{2} }{ \cos^{2} \theta } \\ \\ \\ = \dfrac{(1 - \cos^{2} \theta)(1 + cos^{2} \theta )}{ \cos^{2} \theta} \\ \\ \\ \bf \{ \because {x}^{2} - {y}^{2} = (x - y)(x + y) \} \\ \\ \\ = \dfrac{ \sin^{2} \theta (1 + \cos^{2} \theta)}{ \cos^{2} \theta} \\ \\ \\ \bf \{ \because 1 - \cos^{2} \theta = \sin^{2} \theta \} \\ \\ \\ = \sin^{2} \theta \bigg(\dfrac{1 + cos^{2} \theta}{ \cos^{2} \theta} \bigg) \\ \\ \\ = \sin^{2} \theta \bigg( \dfrac{1}{ \cos^{2} \theta } + \dfrac{ \cos^{2} \theta}{ \cos^{2} \theta} \bigg) \\ \\ \\ = \sin^{2} \theta( \sec^{2} \theta + 1) \\ \\ \\ \{ \because \dfrac{1}{ \cos \theta} = \sec \theta \} \\ \\ \\ = RHS \\ \\ \bf{Hence \ proved }$

Answer 2

To prove:

sec²a-cos²a=sin(sec²a+1)

Explanation:

consider LHS

$\longrightarrow \: \sf \: {sec}^{2} a - \ {cos}^{2} a \\ \\ \longrightarrow \: \sf \: \dfrac{1}{ {cos}^{2} a } - {cos}^{2} a$

$\boxed {\sf since\: \dfrac{1}{cosa}=seca}$

$\longrightarrow \: \sf \: \dfrac{ {1}^{2} - {({cos}^{2}a)}^{2} }{ {cos}^{2}a }$

we know that

$\boxed{\sf a^2-b^2=(a+b)(a-b)}$

$\longrightarrow \: \sf \: \dfrac{(1 + {cos}^{2} a)(1 - {cos}^{2} a)}{ {cos}^{2}a }$

$\boxed{\sf 1-cos^2a=sin^2a}$

$\longrightarrow \: \sf \: {sin}^{2}a( \dfrac{1 + {cos}^{2} a}{ {cos}^{2}a } )$

$\longrightarrow \: \sf \: {sin}^{2} a( \dfrac{1}{ {cos}^{2} a} + \dfrac{ {cos}^{2}a }{ {cos}^{2} a} )$

$\longrightarrow \: \sf \: {sin}^{2}a( {sec}^{2} a + 1)$

$\huge\underline{\sf LHS=RHS}$

Hence proved!!