Question

(secA -cosA) (cota + tan A) - tam A sect.​

Answer 1

Question :

$\\ \bf( sec A - cos A)( cot A + \tan A) = ( tan A. sec A)$

Solution :

In the equation ,

• $\bf{(secA-cosA)(cotA+tanA)=LHS}$

• $\bf{tanAsecA=RHS}$

First let us solve LHS ,

We know that ,

• $\bf{\dfrac{1}{cosA}=secA}$

• $\bf{\dfrac{1}{tanA}=cotA}$

$\\ \longrightarrow \bf \bigg( \frac{1 - {cos}^{2}A }{cosA} \bigg) \bigg( \frac{1 - {tan}^{2}A }{tanA} \bigg)$

Taking LCM ;

$\\ \longrightarrow \bf \bigg( \frac{1 - {cos}^{2} A}{cosA} \bigg) \bigg( \frac{1 + {tan}^{2}A }{tanA} \bigg)$

We know that ,

• $\bf{1-{cos}^ 2A={sin}^2A}$

• $\bf{1+{tan}^2 A= {sec}^2 A}$

$\\ \longrightarrow \bf \bigg( \frac{ {sin}^{2} A}{cosA} \bigg) \bigg( \frac{ {sec}^{2} A}{tanA}\bigg)$

Now ,

• $\bf{tanA=\dfrac{sinA}{cosA}}$

• $\bf{{sec}^2 A=\dfrac{1}{{cos}^2 A}}$

$\\ \longrightarrow \bf \bigg( \frac{sin {}^{2} A}{cosA} \bigg) \bigg( \frac{ \frac{ 1}{cos {}^{2}A } }{ \frac{sinA}{cosA} } \bigg)$

$\\ \longrightarrow \bf \bigg( \frac{{sin}^{2}A }{cosA} \bigg)\bigg( \frac{ 1}{ {cos}^{2}A } \times \frac{cosA}{sinA} \bigg)$

$\\ \longrightarrow \bf \: \frac{sinA}{cosA} \times \frac{1}{cosA}$

$\\ \longrightarrow \bf \: tanA.secA$

$\\ \longrightarrow \bf \: RHS$

Answer 2

Answer:

Question :

\begin{gathered} \\ \bf( sec A - cos A)( cot A + \tan A) = ( tan A. sec A) \\ \end{gathered}

(secA−cosA)(cotA+tanA)=(tanA.secA)

Solution :

In the equation ,

\bf{(secA-cosA)(cotA+tanA)=LHS}(secA−cosA)(cotA+tanA)=LHS

\bf{tanAsecA=RHS}tanAsecA=RHS

First let us solve LHS ,

We know that ,

\bf{\dfrac{1}{cosA}=secA}

cosA

1

=secA

\bf{\dfrac{1}{tanA}=cotA}

tanA

1

=cotA

\begin{gathered} \\ \longrightarrow \bf \bigg( \frac{1 - {cos}^{2}A }{cosA} \bigg) \bigg( \frac{1 - {tan}^{2}A }{tanA} \bigg) \\ \\ \end{gathered}

⟶(

cosA

1−cos

2

A

)(

tanA

1−tan

2

A

)

Taking LCM ;

\begin{gathered} \\ \longrightarrow \bf \bigg( \frac{1 - {cos}^{2} A}{cosA} \bigg) \bigg( \frac{1 + {tan}^{2}A }{tanA} \bigg) \\ \end{gathered}

⟶(

cosA

1−cos

2

A

)(

tanA

1+tan

2

A

)

We know that ,

\bf{1-{cos}^ 2A={sin}^2A}1−cos

2

A=sin

2

A

\bf{1+{tan}^2 A= {sec}^2 A}1+tan

2

A=sec

2

A

\begin{gathered} \\ \longrightarrow \bf \bigg( \frac{ {sin}^{2} A}{cosA} \bigg) \bigg( \frac{ {sec}^{2} A}{tanA}\bigg) \\ \\ \end{gathered}

⟶(

cosA

sin

2

A

)(

tanA

sec

2

A

)

Now ,

\bf{tanA=\dfrac{sinA}{cosA}}tanA=

cosA

sinA

\bf{{sec}^2 A=\dfrac{1}{{cos}^2 A}}sec

2

A=

cos

2

A

1

\begin{gathered} \\ \longrightarrow \bf \bigg( \frac{sin {}^{2} A}{cosA} \bigg) \bigg( \frac{ \frac{ 1}{cos {}^{2}A } }{ \frac{sinA}{cosA} } \bigg) \\ \\ \end{gathered}

⟶(

cosA

sin

2

A

)(

cosA

sinA

cos

2

A

1

)

\begin{gathered} \\ \longrightarrow \bf \bigg( \frac{{sin}^{2}A }{cosA} \bigg)\bigg( \frac{ 1}{ {cos}^{2}A } \times \frac{cosA}{sinA} \bigg) \\ \\ \end{gathered}

⟶(

cosA

sin

2

A

)(

cos

2

A

1

×

sinA

cosA

)

\begin{gathered} \\ \longrightarrow \bf \: \frac{sinA}{cosA} \times \frac{1}{cosA} \\ \\ \end{gathered}

⟶

cosA

sinA

×

cosA

1

\begin{gathered} \\ \longrightarrow \bf \: tanA.secA \\ \\ \end{gathered}

⟶tanA.secA

\begin{gathered} \\ \longrightarrow \bf \: RHS\end{gathered}

⟶RHS