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Trigonometric Identities Final Assessment

Section A: Identity Concepts

Question 1mcq
Which statement is true about an identity?
A.It is true for all allowed angle values.
B.It is true for exactly one angle.
C.It must contain no fractions.
D.It is always false until solved.
Question 2mcq
Which step correctly starts a proof of sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1 from a right triangle?
A.Write sin⁡A=ho\sin A=\frac{h}{o}.
B.Assume sin⁡A+cos⁡A=1\sin A+\cos A=1.
C.Use o+a=ho+a=h.
D.Write sin⁡A=oh\sin A=\frac{o}{h} and cos⁡A=ah\cos A=\frac{a}{h}.
Question 3mcq
If cos⁡A=725\cos A=\frac{7}{25} for acute AA, find sin⁡A\sin A.
A.724\frac{7}{24}
B.2524\frac{25}{24}
C.2425\frac{24}{25}
D.1825\frac{18}{25}
Question 4mcq
Which identity follows by dividing sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1 by cos⁡2A\cos^2A?
A.sec⁡2A+1=tan⁡2A\sec^2A+1=\tan^2A
B.tan⁡2A+1=sec⁡2A\tan^2A+1=\sec^2A
C.cot⁡2A+1=cosec⁡2A\cot^2A+1=\cosec^2A
D.sin⁡2A+1=cos⁡2A\sin^2A+1=\cos^2A
Question 5mcq
Which identity follows by dividing sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1 by sin⁡2A\sin^2A?
A.1+cot⁡2A=cosec⁡2A1+\cot^2A=\cosec^2A
B.1+tan⁡2A=sec⁡2A1+\tan^2A=\sec^2A
C.sin⁡2A+1=cos⁡2A\sin^2A+1=\cos^2A
D.cosec⁡2A+1=cot⁡2A\cosec^2A+1=\cot^2A
Question 6mcq
Simplify 1−sin⁡2A1-\sin^2A.
A.sin⁡2A\sin^2A
B.tan⁡2A\tan^2A
C.1+cos⁡2A1+\cos^2A
D.cos⁡2A\cos^2A

Section B: Simplification and Proof

Question 7mcq
Which expression is equal to 1sec⁡2A\frac{1}{\sec^2A}?
A.tan⁡2A\tan^2A
B.cot⁡2A\cot^2A
C.cos⁡2A\cos^2A
D.sin⁡2A\sin^2A
Question 8mcq
Which replacement helps prove identities involving cosec⁡A\cosec A?
A.cosec⁡A=cos⁡A\cosec A=\cos A
B.cosec⁡A=1sin⁡A\cosec A=\frac1{\sin A}
C.cosec⁡A=1cos⁡A\cosec A=\frac1{\cos A}
D.cosec⁡A=1tan⁡A\cosec A=\frac1{\tan A}
Question 9mcq
Simplify (sec⁡A−tan⁡A)(sec⁡A+tan⁡A)(\sec A-\tan A)(\sec A+\tan A).
A.11
B.sec⁡2A+tan⁡2A\sec^2A+\tan^2A
C.tan⁡2A\tan^2A
D.sec⁡A\sec A
Question 10mcq
If cot⁡A=512\cot A=\frac{5}{12}, find cosec⁡2A\cosec^2A.
A.25144\frac{25}{144}
B.144169\frac{144}{169}
C.1312\frac{13}{12}
D.169144\frac{169}{144}
Question 11mcq
Which condition is needed before dividing by sin⁡A\sin A?
A.cos⁡A=1\cos A=1
B.tan⁡A=0\tan A=0
C.sin⁡Aeq0\sin A eq0
D.sin⁡A=0\sin A=0
Question 12mcq
A student proves an identity by checking only A=30∘A=30^\circ. Why is this not enough?
A.Only 45∘45^\circ can be used in identities.
B.An identity must be shown for all allowed values, not one example.
C.30∘30^\circ values are never allowed.
D.Checking one angle proves every identity.

Section C: Evaluation and Restrictions

Question 13mcq
If sin⁡A=12\sin A=\frac{1}{2} for acute AA, find cos⁡2A\cos^2A.
A.34\frac{3}{4}
B.14\frac{1}{4}
C.12\frac{1}{2}
D.32\frac{\sqrt3}{2}
Question 14mcq
Simplify cosec⁡2A−1\cosec^2A-1.
A.tan⁡2A\tan^2A
B.sec⁡2A\sec^2A
C.sin⁡2A\sin^2A
D.cot⁡2A\cot^2A
Question 15mcq
To prove 11+tan⁡2A=cos⁡2A\frac{1}{1+\tan^2A}=\cos^2A, which substitution is best?
A.tan⁡A=cos⁡A\tan A=\cos A
B.sec⁡2A=sin⁡2A\sec^2A=\sin^2A
C.1+tan⁡2A=sec⁡2A1+\tan^2A=\sec^2A
D.1+tan⁡2A=cosec⁡2A1+\tan^2A=\cosec^2A
Question 16mcq
Simplify cos⁡A⋅sec⁡A\cos A\cdot\sec A where defined.
A.00
B.11
C.cos⁡2A\cos^2A
D.sec⁡2A\sec^2A
Question 17mcq
Which factorisation is correct?
A.sec⁡2A−1=(sec⁡A−1)(sec⁡A+1)\sec^2A-1=(\sec A-1)(\sec A+1)
B.sec⁡2A−1=(sec⁡A−1)2\sec^2A-1=(\sec A-1)^2
C.sec⁡2A−1=sec⁡A2\sec^2A-1=\sec A^2
D.sec⁡2A−1=(tan⁡A−1)2\sec^2A-1=(\tan A-1)^2
Question 18mcq
If sec⁡A=1312\sec A=\frac{13}{12}, find tan⁡2A\tan^2A.
A.169144\frac{169}{144}
B.14425\frac{144}{25}
C.512\frac{5}{12}
D.25144\frac{25}{144}

Section D: Misconception Check

Question 19mcq
For which expression must cos⁡Aeq0\cos A eq0?
A.sin⁡A\sin A
B.cos⁡2A\cos^2A
C.tan⁡A=sin⁡Acos⁡A\tan A=\frac{\sin A}{\cos A}
D.sin⁡2A+cos⁡2A\sin^2A+\cos^2A
Question 20mcq
A student changes sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1 to sin⁡A+cos⁡A=1\sin A+\cos A=1. What is wrong?
A.The equation becomes true after adding tangent.
B.Square terms cannot be dropped.
C.It is correct for all angles.
D.Only cosine should be dropped.
Question 21mcq
Which one is an equation that may be true only for some angles, not an identity?
A.sin⁡A=12\sin A=\frac12
B.sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1
C.1+tan⁡2A=sec⁡2A1+\tan^2A=\sec^2A
D.1+cot⁡2A=cosec⁡2A1+\cot^2A=\cosec^2A
Question 22mcq
In the right-triangle derivation, why does o2+a2=h2o^2+a^2=h^2?
A.By equal tangents theorem
B.By midpoint formula
C.By AP sum formula
D.By Pythagoras theorem
Question 23mcq
If tan⁡2A=916\tan^2A=\frac{9}{16}, then sec⁡2A\sec^2A is:
A.1625\frac{16}{25}
B.34\frac{3}{4}
C.2516\frac{25}{16}
D.716\frac{7}{16}
Question 24mcq
If cot⁡2A=2425\cot^2A=\frac{24}{25}, then cosec⁡2A\cosec^2A is:
A.2549\frac{25}{49}
B.4925\frac{49}{25}
C.2425\frac{24}{25}
D.125\frac{1}{25}