Question

u can sol it trignometry​

Answer 1

Answers:

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

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

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

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

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

>>WE TAKE A RIGHT-ANGLED TRIANGLE

WHERE /_A AND /_C ARE ACUTE ANGLES AND /_B=90°

WE KNOW THAT,

$\cos(A) = \frac{1}{ \sec(A) } \\ also \: \\ { \sin(A) }^{2} + { \cos(A) }^{2} = 1 \\ { \sin(A) }^{2} = 1 - { \cos(A) }^{2} \\ \sin(A) = \sqrt{1 - ( \frac{1}{ \sec(A) })^{2} } \\ \sin(A) = \sqrt{ \frac{ \sec^{2}(A) - 1 }{ \sec ^{2} (A) } }=\frac{\sqrt{\sec^{2}(A)-1}}{\sec(A)}\\\\tan^{2}(A)+1=sec^{2}(A)\\tan^{2}=sec^{2}(A)-1\\tan(A)=\sqrt{sec^{2}(A)-1}\\\\cot(A)=\frac{cos(A)}{sin(A)}=\frac{\frac{1}{sec(A)}}{\frac{\sqrt{sec^{2}(A)-1}}{sec(A)}}\\cot(A)=\frac{1}{\sqrt{sec^{2}(A)-1}}\\\\cosec(A)=\frac{1}{sin(A)}=\frac{sec(A)}{\sqrt{sec^{2}(A)-1}}\\cosec(A)=\frac{sec(A)}{\sqrt{sec^{2}(A)-1}}$