Fundamental Trigonometric Identities

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

1. CSCA Math December 2025 Past Paper
   $\left(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\right)\tan\frac{\pi}{4}=$ ( ).
   • A．$2\sqrt{2}$
   • B．1
   • C．$\sqrt{2}$
   • D．2
   Show answer & solution

   Answer: C

   Answer: C $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ and $\tan\frac{\pi}{4}=1$, so the value is $\sqrt{2}$. Choose C.
   See it in the full CSCA Math December 2025 Past Paper
2. CSCA Math January 2026 Past Paper
   Given that $\sin \alpha = \frac { 4 } { 5 }$, and the angle $\alpha$ belongs to the second quadrant, then which of the following statements is correct? ()
   • A．$\cos \alpha = \frac { 3 } { 5 }$
   • B．$\cos \alpha = - \frac { 3 } { 5 }$
   • C．$\tan \alpha = - \frac { 3 } { 4 }$
   • D．$\tan \alpha = \frac { 4 } { 3 }$
   Show answer & solution

   Answer: B

   Given $\sin \alpha = 4 / 5$ and $\alpha$ is in Quadrant II. Find $\cos \alpha$. In Quadrant II, sine is positive and cosine is negative. Use $\sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha = 1 : \cos ^ { 2 } \alpha = 1 - ( 4 / 5 ) ^ { 2 } = 1 - 16 / 25 = 9 / 25$. So $\cos \alpha = - 3 / 5$.
   Take the full CSCA Math January 2026 Past Paper — timed & graded
3. CSCA Math April 2026 Past Paper
   If point $P ( x , y )$ lies on the terminal side of angle $\alpha$ with $r = \sqrt { x ^ { 2 } + y ^ { 2 } }$, then $\cos \alpha =$
   • A．$\frac { y } { r }$
   • B．$\frac { y } { x }$
   • C．$\frac { x } { y }$
   • D．$\frac { x } { r }$
   Show answer & solution

   Answer: D

   By definition $\cos \alpha = \frac { x } { r }$. (Note: $\frac { y } { r } = \sin \alpha , \frac { y } { x } = \tan \alpha$.)
   Take the full CSCA Math April 2026 Past Paper — timed & graded