Inverse trigonometric functions answer one simple question: given a ratio, what is the angle? If sinθ = ½, then θ = sin−1(½). For NDA, this chapter is a steady source of 1–2 marks every year — mostly about principal values, domains, ranges and a handful of identities. Get those right and the questions become almost automatic.
Why Inverse Trig Matters in NDA
Ordinary trigonometry takes an angle and gives a ratio: sin 30° = ½. Inverse trigonometry reverses this — it takes a ratio and returns the angle. The symbol sin−1x is read as “angle whose sine is x”, not as 1÷sin x. Some books write this same function as arcsin x, arccos x and so on, so do not get confused if you see that notation in a question; it means exactly the same thing.
Why bother reversing? Because real problems often hand you a ratio and ask for the angle. A ladder leaning against a wall, the bearing of an aircraft, the slope of a road — in each case you know the sides and want the angle. Inverse trigonometric functions are the tool that turns a known ratio back into the angle that created it.
In the NDA written exam, this topic appears almost every year inside the trigonometry block. The questions are usually direct and formula-based — finding a principal value, simplifying an expression, or solving a short equation. Rarely will you face a long multi-step proof. That makes it one of the highest reward-per-effort topics in the entire Maths paper, and a smart student treats it as a guaranteed mark rather than a gamble.
sin−1x is not (sin x)−1. The −1 here means “inverse function”, not reciprocal. The reciprocal of sin x is cosec x, written (sin x)−1. The exponent and the inverse symbol look identical, so always read the position carefully.
Why We Restrict the Domain
A function can have an inverse only if it is one-one (each output comes from exactly one input) and onto. But sin x repeats its values endlessly — sin 0 = sin π = sin 2π = 0. So sine is not one-one over all real numbers, and without a fix it would have no proper inverse at all.
To fix this, we shrink the input to a single stretch where the function rises or falls without repeating. For sine we choose [−π/2, π/2]. On this interval sine increases steadily from −1 to 1, so each value of sine matches exactly one angle. That chosen interval becomes the range of sin−1x, also called the principal value branch. Every inverse trig function has its own such branch, decided by international convention so that everyone gets the same answer.
You could, in theory, have picked a different interval where sine is also one-one, such as [π/2, 3π/2]. Mathematicians simply agreed on the interval closest to zero, which keeps the answers small and tidy. That is why these particular ranges are fixed and must be memorised exactly.
The domain of an inverse trig function is the set of ratios you can plug in. Its range is the set of angles you can get out. Mixing these two up is the single biggest cause of wrong answers in this chapter.
Domains and Ranges of All Six Functions
This table is the heart of the chapter. If you memorise nothing else, memorise this. The principal value of any inverse trig function must lie inside its listed range.
- sin−1x: domain [−1, 1], range [−π/2, π/2]
- cos−1x: domain [−1, 1], range [0, π]
- tan−1x: domain all real numbers, range (−π/2, π/2)
- cot−1x: domain all real numbers, range (0, π)
- sec−1x: domain |x| ≥ 1, range [0, π] − {π/2}
- cosec−1x: domain |x| ≥ 1, range [−π/2, π/2] − {0}
Notice the pattern: sin−1, tan−1 and cosec−1 all use ranges centred on the fourth-and-first quadrant [−π/2, π/2], while cos−1, cot−1 and sec−1 use the first-and-second quadrant [0, π].
For functions with range [−π/2, π/2], the answer for a negative input is a negative angle. For functions with range [0, π], a negative input gives an angle in the second quadrant (greater than π/2). This one rule decides most sign-based traps.
Finding the Principal Value
The principal value is the unique angle inside the range that produces the given ratio. Every other angle that gives the same ratio is ignored; only the one inside the principal branch is accepted. To find it, follow three steps:
- Identify which function you are dealing with and recall its range.
- Find the basic angle (ignore the sign for a moment) whose ratio matches the number.
- Adjust the sign so the final angle lands inside the correct range.
For example, cos−1(−½): the basic angle for ½ is π/3. Since cos−1 has range [0, π] and the input is negative, the answer sits in the second quadrant: π − π/3 = 2π/3.
It helps to keep a short mental list of standard angles and their ratios: 0, π/6, π/4, π/3 and π/2 cover almost every value the NDA paper throws at you. If you know that tan (π/4) = 1, sin (π/6) = ½ and cos (π/6) = √3/2, you can read off most principal values in seconds without any calculation.
Students often write sin−1(−½) = 7π/6 because sin (7π/6) = −½. But 7π/6 is outside the range [−π/2, π/2]. The correct principal value is −π/6.
Identities for Negative Arguments
These six identities tell you how each function behaves when the input becomes negative. They flow directly from the chosen ranges.
sin−1(−x) = −sin−1x
tan−1(−x) = −tan−1x
cosec−1(−x) = −cosec−1x
cos−1(−x) = π − cos−1x
cot−1(−x) = π − cot−1x
sec−1(−x) = π − sec−1x
The first group (sine type) is odd — the minus simply moves outside. The second group (cosine type) reflects about π. This split matches the range pattern you saw earlier, so it is easy to recall. To check your memory, test a known value: cos−1(−1) should equal π − cos−1(1) = π − 0 = π, which is correct because cos π = −1. A quick sanity check like this catches a wrong sign before it costs you a mark.
Reciprocal and Complementary Identities
Two more sets of identities make simplification fast. The reciprocal identities connect a function to its reciprocal partner:
cosec−1x = sin−1(1/x)
sec−1x = cos−1(1/x)
cot−1x = tan−1(1/x), for x > 0
The complementary identities pair each function with its co-function. Their sum is always π/2:
sin−1x + cos−1x = π/2
tan−1x + cot−1x = π/2
sec−1x + cosec−1x = π/2
If a question gives you sin−1x and asks about cos−1x of the same number, you almost never need to compute both. Just subtract from π/2. This saves real time on the clock.
Addition and Subtraction Formulas
The tan−1 addition formula is the most tested in NDA, so learn it cold.
tan−1x + tan−1y = tan−1[(x + y)/(1 − xy)], when xy < 1
tan−1x − tan−1y = tan−1[(x − y)/(1 + xy)]
If xy > 1 (with both x and y positive), add π to the result, because the true sum then crosses π/2. Two more useful double-angle results:
2 tan−1x = tan−1[2x/(1 − x2)] = sin−1[2x/(1 + x2)] = cos−1[(1 − x2)/(1 + x2)]
Forgetting the condition xy < 1. If you ignore it you may report a value outside the correct range — a classic NDA trap where the “obvious” option is wrong.
Worked Example
Let us combine several ideas in one problem of the kind NDA loves.
Evaluate tan−1(1) + cos−1(−½) + sin−1(−½).
Notice how each term was forced into its own correct range before adding. The negative input in the cosine term produced a second-quadrant angle, while the negative input in the sine term produced a negative angle. That is the whole skill in one problem.
Graphs and Behaviour at a Glance
You do not need to draw perfect graphs, but knowing the shape prevents range errors.
- sin−1x rises from −π/2 at x = −1 to π/2 at x = 1, passing through the origin.
- cos−1x falls from π at x = −1 to 0 at x = 1. It is the only one of the basic three that decreases.
- tan−1x rises across all real x but never reaches π/2 or −π/2 — those are horizontal asymptotes.
As x → +∞, tan−1x → π/2 and as x → −∞, tan−1x → −π/2. The function gets close but never touches these limits.
One more visual fact worth remembering: the graph of any inverse function is the mirror image of the original function across the line y = x. So the curve of sin−1x is simply the restricted sine curve flipped over that diagonal. Picturing this reflection helps you instantly recall where each inverse graph starts and ends.
Previous-Year Style Question
Here is a question in the exact format you can expect in the NDA paper, with a clean solution.
Q. The value of tan−1(1/2) + tan−1(1/3) is equal to: (a) π/6 (b) π/4 (c) π/3 (d) π/2
Answer: (b) π/4. Here x = 1/2, y = 1/3, so xy = 1/6 < 1. Apply the formula: tan−1[(1/2 + 1/3)/(1 − 1/6)] = tan−1[(5/6)/(5/6)] = tan−1(1) = π/4.
The trick is always to check xy < 1 first, then substitute carefully. Because xy was clearly less than 1, no π adjustment was needed. A neat shortcut for the exam hall: whenever the two numerators and the resulting fraction collapse to 1, the answer is π/4 — a pattern that shows up often in NDA papers and lets you mark the option almost on sight.
Common Traps to Avoid
Examiners design options to catch predictable slips. Watch for these:
- Treating sin−1x as a reciprocal — it is an inverse function.
- Giving an answer outside the principal range, especially for negative inputs.
- Using the tan−1 addition formula without checking xy < 1.
- Confusing the range [0, π] of cos−1 with the range [−π/2, π/2] of sin−1.
- Plugging |x| < 1 into sec−1 or cosec−1, where it is undefined.
Before marking any answer, ask yourself one question: does my angle lie inside the function's range? If not, fix the sign or shift by π until it does.
Quick Revision
- sin−1x: domain [−1,1], range [−π/2, π/2]; cos−1x: range [0, π].
- tan−1x: domain all reals, range (−π/2, π/2).
- sin−1x + cos−1x = π/2, and similarly for the other co-pairs.
- tan−1x + tan−1y = tan−1[(x+y)/(1−xy)] when xy < 1.
- Negative input: sine type gives a negative angle; cosine type gives π − angle.
- Always confirm the final angle lies inside the principal range.
Drill the domain-range table until you can recite it instantly. In the exam, recognising the correct range is usually most of the work — the arithmetic that follows is short.
Frequently asked questions
Is sin inverse the same as 1 over sin?
No. sin−¹x means the angle whose sine is x, an inverse function. The reciprocal of sin x is cosec x, which is (sin x)−¹. Confusing the two is the most common error in this topic.
What is a principal value in inverse trigonometry?
It is the single angle, lying inside the function's fixed range, that produces the given ratio. For example, the principal value of sin−¹(½) is π/6, even though many angles have sine equal to ½.
Why does cos inverse have range 0 to pi but sin inverse has minus pi by 2 to pi by 2?
Each range is chosen as an interval where that function is one-one and covers all its values. Cosine is one-one over [0, π], while sine is one-one over [−π/2, π/2]. These choices are fixed conventions.
How important is this topic for the NDA exam?
It is consistently worth about 1 to 2 marks every year and the questions are usually direct. Because the effort needed is low, The Cavalier rates it among the best scoring topics in NDA Maths trigonometry.
When do I add pi to the tan inverse addition formula?
When both x and y are positive and their product xy is greater than 1. In that case the true sum of the two angles exceeds π/2, so you add π to the value the formula gives.
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