JEE Maths›Trigonometric Functions›Visual Solution
Visual SolutionTricky
Question
The value of tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ is:
(A)44
(B)22
(C)11
(D)00
Solution Path
Complementary angles give (tan⁡+cot⁡)(\tan + \cot) pairs, use tan⁡x+cot⁡x=2/sin⁡2x\tan x + \cot x = 2/\sin 2x, substitute sin⁡18∘\sin 18^\circ and sin⁡54∘\sin 54^\circ to get 4.
01Question Setup
1/5
Find the value of tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ.
tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘=  ?\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = \;?
02Complementary Angle Regrouping
2/5
Recognize tan⁡81∘=cot⁡9∘\tan 81^\circ = \cot 9^\circ and tan⁡63∘=cot⁡27∘\tan 63^\circ = \cot 27^\circ. Regroup as (tan⁡9∘+cot⁡9∘)−(tan⁡27∘+cot⁡27∘)(\tan 9^\circ + \cot 9^\circ) - (\tan 27^\circ + \cot 27^\circ).
(tan⁡θ+cot⁡θ)(\tan\theta + \cot\theta) pairs
03Key IdentityKEY INSIGHT
3/5
tan⁡x+cot⁡x=sin⁡xcos⁡x+cos⁡xsin⁡x=1sin⁡xcos⁡x=2sin⁡2x\tan x + \cot x = \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{1}{\sin x \cos x} = \dfrac{2}{\sin 2x}. So the expression becomes 2sin⁡18∘−2sin⁡54∘\dfrac{2}{\sin 18^\circ} - \dfrac{2}{\sin 54^\circ}.
tan⁡x+cot⁡x=2sin⁡2x\tan x + \cot x = \dfrac{2}{\sin 2x}
04Substitute Known Values
4/5
sin⁡18∘=5−14\sin 18^\circ = \dfrac{\sqrt{5}-1}{4}, sin⁡54∘=5+14\sin 54^\circ = \dfrac{\sqrt{5}+1}{4}. Compute: 85−1−85+1=8⋅24=4\dfrac{8}{\sqrt{5}-1} - \dfrac{8}{\sqrt{5}+1} = 8 \cdot \dfrac{2}{4} = 4.
=4= 4
05Final Answer
5/5
tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘=4\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = 4. Answer is (A).
4\boxed{4}
Concepts from this question3 concepts unlocked
★

Compound Angle Expansion

EASY

Expand sin(A+B) and cos(A+B) into products of individual trig ratios of A and B

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B

The foundation of nearly all trig manipulation in JEE. Double angle, half angle, and product formulas all derive from this. Appears in 3-4 questions every paper.

Double/half angle derivationsTrig simplificationProving identities
Practice (15 Qs) →

Product-to-Sum and Sum-to-Product

STANDARD

Convert products of sin/cos into sums (and vice versa) to simplify expressions or solve equations

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A+B) + \sin(A-B)

Turns complicated products into manageable sums. Essential for integrating trig products and solving equations where factoring is needed.

Trig simplificationIntegration of trig productsSeries summation
Practice (10 Qs) →

Conditional Identities (A+B+C = pi)

TRICKY

Special trig identities that hold when angles sum to pi, commonly used in triangle-related problems

tan⁡A+tan⁡B+tan⁡C=tan⁡A tan⁡B tan⁡C\tan A + \tan B + \tan C = \tan A\,\tan B\,\tan C

Appears in both trig and properties-of-triangle chapters. JEE loves asking prove-that or simplify problems under the condition A+B+C = pi.

Triangle identitiesProve-that problemsMulti-angle simplification
Practice (8 Qs) →
Want more practice?
Try more PYQs from this chapter →