Math, asked by vaishaliranpura, 1 year ago

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

Answers

Answered by Anonymous
8

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 }

Answered by DhanyaDA
6

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!!

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