Question

Express sin A in terms of sec A and tan A. please answer it as soon as possible

Answer 1

Question :-

Express sin A in terms of sec A and tan A

Answer :-

Required to find :-

Sin A in the terms of sec A and tan A ?

Identity used :-

Sin² θ + Cos² θ = 1

Solution :-

We need to express Sin A in the terms of sec A and tan A .

So,

Consider the identity ;

Sin²θ + cos²θ = 1

First, let's express sin θ in terms of cos θ

This implies ;

Here,

θ ( theta ) = A

So, the identity becomes ;

sin² A + cos² A = 1

since, we want to express sin A in terms of cos . Transpose sin² A to the right side

cos² A = 1 - sin² A

cos A = √1 - sin² A

Hence,

Sin A is expressed in terms of cos A as ; √1 - sin² A

Now, using this idea let's express Sin A in terms of sec A , tan A .

So,

we know that ;

sec A = 1/ cos A

( since, cos A = √1 - sin² A )

This implies,

sec A = 1/ √1 - sin² A

Hence,

Sin A in terms of sec A = 1/√ 1 - sin² A

Similarly,

Now, let's find how sin A is expressed in terms of tan A

Since we know that ;

Tan A = sin A/cos A

This implies ;

( since, cos A = √1 - sin² A )

Tan A = sin A / √1 - sin² A

Hence,

Sin A in terms of tan A = sin A/ √1 - sin² A

Therefore,

$\boxed{ \tt{ \sec A = \frac{1}{ \sqrt{1 - { \sin}^{2} A } } }}$

$\boxed{ \tt{ \tan A = \dfrac{ \sin A}{ \sqrt{1 - { \sin}^{2} A } }}}$

Answer 2

$\dag \: { \underline{ \red{ \tt{SoluTion :-}}}}$

${\tt{ As \: \: we \:\: know \:\: that, }}$

$\tt \longrightarrow \: sin {}^{2} A + cos {}^{2} A = 1 \\\\ \tt \green{[ Where, \: \: \theta \: \: is \: \: A. ]} \\ \\ \tt \longrightarrow .°. \: cos {}^{2} A = 1 - sin {}^{2} A \\ \\ \tt \longrightarrow .°. \: \purple{cosA = \sqrt{1 - sin^2A}}\\\\ \tt \green{[ °.° secA = \frac{1}{cosA} ]} \\ \\\tt \longrightarrow \: \red{\boxed{\tt{secA = \frac{1}{ \sqrt{1 - sin {}^{2} A} }}}} \\ \\\tt \green{ [ °.° \: tanA = \frac{sinA}{cosA} ]} \\\\ \tt \green{[ \:Where, \: \: cosA \: \: = \: \: { \sqrt{1 - sin {}^{2} A} } \: ] } \\ \\\tt \longrightarrow \: \red{\boxed{\tt{tanA \: = \frac{sinA}{ \sqrt{1 - sin {}^{2} A} }}} }$