Question

hey guyz
answer this plz
​

Answer 1

We are asked to write all other trigonometric ratios of the angle A in terms of $\sf{\sec A.}$

Since $\sf{\sec A}$ and $\sf{\cos A}$ are reciprocals to each other,

$\longrightarrow\sf{\underline{\underline{\cos A=\dfrac{1}{\sec A}}}}$

We know the Pythagorean Trigonometric Identities,

$\longrightarrow\sf{\sec^2A-\tan^2A=1\quad\quad\dots(1)}$

$\longrightarrow\sf{\sin^2A+\cos^2A=1\quad\quad\dots(2)}$

From (1),

$\longrightarrow\sf{\tan^2A=\sec^2A-1}$

$\longrightarrow\sf{\underline{\underline{\tan A=\sqrt{\sec^2A-1}}}}$

On taking reciprocal of this equation,

$\longrightarrow\sf{\underline{\underline{\cot A=\dfrac{1}{\sqrt{\sec^2A-1}}}}}$

From (2),

$\longrightarrow\sf{\sin^2A=1-\cos^2A}$

$\longrightarrow\sf{\sin^2A=1-\dfrac{1}{\sec^2A}}$

$\longrightarrow\sf{\sin^2A=\dfrac{\sec^2A-1}{\sec^2A}}$

$\longrightarrow\sf{\underline{\underline{\sin A=\dfrac{\sqrt{\sec^2A-1}}{\sec A}}}}$

On taking the reciprocal of this equation,

$\longrightarrow\sf{\underline{\underline{\csc A=\dfrac{\sec A}{\sqrt{\sec^2A-1}}}}}$

Finally,

• $\sf{\underline{\underline{\sin A=\dfrac{\sqrt{\sec^2A-1}}{\sec A}}}}$

• $\sf{\underline{\underline{\cos A=\dfrac{1}{\sec A}}}}$

• $\sf{\underline{\underline{\tan A=\sqrt{\sec^2A-1}}}}$

• $\sf{\underline{\underline{\csc A=\dfrac{\sec A}{\sqrt{\sec^2A-1}}}}}$

• $\sf{\underline{\underline{\cot A=\dfrac{1}{\sqrt{\sec^2A-1}}}}}$

Answer 2

$\sf{To\:Write \: angle\:in \:term\:of\:secA}$

In Cos A

• $\boxed{\sf{ \cos A = \frac{1}{ \sec A} }}$

In sin A

$\sf{ \sin A = \sqrt{ 1 - {\cos}^{2} A }}$

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

• $\boxed{\sf{ \sin(a) = \frac{ \sqrt{ { \sec(a)^{2} - 1 }^{2} } }{ \sec(a) } }}$

In tan A

$\sf{{\tan}^{2} A = {\sec}^{2} A - 1}$

• $\boxed {\sf{ \tan(a) = \sqrt{ { \sec(a) }^{2} - 1} }}$

In Cot A

Cot is the reciprocal of tan A . So ,

• $\boxed{ \sf{ \cot(a) = \frac{1}{ \sqrt{ { \sec(a) }^{2} - 1 } } }}$

In Cosec A

Cosec is the reciprocal of sin A .

$\sf{ \sin(a) = \frac{ \sqrt{ { \sec(a)^{2} - 1 }^{2} } }{ \sec(a) } }$

It's reciprocal value is equal to Cosec A

• $\boxed{\sf{\cosec A = \frac{\sec A}{\sqrt{{\sec}^{2} A - 1}}}}$

In sec A

• Sec A = Sec A.

So we converted all trigonometric ratios in term of sec A .

Some identities to remember .

${ \sin(a) }^{2} + { \cos(a) }^{2} = 1$

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

${ \cot(a) }^{2} + 1 = { \csc(a) }^{2}$