Math IOQM

  Statement of the identity cot ⁡  ⁣ ( θ 2 ) = 1 + cos ⁡ θ sin ⁡ θ \boxed{\displaystyle \cot\!\left(\tfrac{\theta}{2}\right)=\frac{1+\cos\theta}{\sin\theta}} This relates the cotangent of half an angle to the sine and cosine of the whole angle (with θ ≢ 2 k π \theta\not\equiv 2k\pi so sin ⁡ θ ≠ 0 \sin\theta\neq0 ). Proof using double-angle formulas Start from double-angle expressions sin ⁡ θ    =    2 sin ⁡  ⁣ ( θ 2 ) cos ⁡  ⁣ ( θ 2 ) , 1 + cos ⁡ θ    =    2 cos ⁡ 2  ⁣ ( θ 2 ) . \sin\theta \;=\; 2\sin\!\bigl(\tfrac{\theta}{2}\bigr)\cos\!\bigl(\tfrac{\theta}{2}\bigr), \quad 1+\cos\theta \;=\; 2\cos^{2}\!\bigl(\tfrac{\theta}{2}\bigr). Form the ratio 1 + cos ⁡ θ sin ⁡ θ \dfrac{1+\cos\theta}{\sin\theta} 1 + cos ⁡ θ sin ⁡ θ    =    2 cos ⁡ 2  ⁣ ( θ 2 ) 2 sin ⁡  ⁣ ( θ 2 ) cos ⁡  ⁣ ( θ 2 )    =    2 cos ⁡ 2  ⁣ ( θ 2 ) 2 sin ⁡  ⁣ ( θ 2 ) cos ⁡  ⁣ ( θ 2 )    =    cos ⁡  ⁣ ( θ 2 ) sin ⁡  ⁣ ( θ 2 ) . \frac{1+\cos\theta}{\sin\theta} \;=\; \frac{2\cos^{2}\!\bigl(\tfrac{\theta}{2}...

q24 - The sides of a Triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine perimeter of the triangle. Answer: 15 Solution: Let the sides be n , n + 1 , n + 2 n, n+1, n+2 . i.e., A C = n , A B = n + 1 , B C = n + 2 AC = n, AB = n+1, BC = n+2 Smallest angle is B B and largest one is A A . Here ∠ A = 2 ∠ B \angle A = 2 \angle B Also, ∠ A + ∠ B + ∠ C = 180 ∘ \angle A + \angle B + \angle C = 180^\circ ⇒ 3 ∠ B + ∠ C = 180 ∘ ⇒ ∠ C = 180 ∘ − 3 ∠ B \Rightarrow 3 \angle B + \angle C = 180^\circ \Rightarrow \angle C = 180^\circ - 3 \angle B We have, sine law as, sin ⁡ A n + 2 = sin ⁡ B n = sin ⁡ C n + 1 \frac{\sin A}{n+2} = \frac{\sin B}{n} = \frac{\sin C}{n+1} ⇒ sin ⁡ 2 B n + 2 = sin ⁡ B n = sin ⁡ ( 180 ∘ − 3 B ) n + 1 \Rightarrow \frac{\sin 2B}{n+2} = \frac{\sin B}{n} = \frac{\sin (180^\circ - 3B)}{n+1} ⇒ sin ⁡ 2 B n + 2 = sin ⁡ B n = sin ⁡ 3 B n + 1 \Rightarrow \frac{\sin 2B}{n+2} = \frac{\sin B}{n} = \frac{\sin 3B}{n+1} From (i)...

Q25 "Given that ( 1 + sin ⁡ t ) ( 1 + cos ⁡ t ) = 5 4 (1 + \sin t)(1 + \cos t) = \frac{5}{4} and ( 1 − sin ⁡ t ) ( 1 − cos ⁡ t ) = m n − k (1 - \sin t)(1 - \cos t) = \frac{m}{n} - \sqrt{k} , where k , m k, m and n n are positive integers with m m and n n relatively prime, find k + m + n + 3 k + m + n + 3 " Answer: 30 Solution sketch Let S = sin ⁡ t S=\sin t and C = cos ⁡ t C=\cos t . 1. Use the first condition ( 1 + S ) ( 1 + C ) = 5 4 (1+S)(1+C)=\tfrac54 ⟹    1 + S + C + S C = 5 4          ⇒       S + C + S C = 1 4 . (A) \Longrightarrow\;1+S+C+SC=\tfrac54\;\;\; \Rightarrow\;\; S+C+SC=\tfrac14. \tag{A} Define u = S + C , v = S C . u=S+C,\quad v=SC . Then (A) says u + v = 1 4 u+v=\tfrac14 . 2. Relate u u and v v with S 2 + C 2 = 1 S^2+C^2=1 S 2 + C 2 = ( S + C ) 2 − 2 S C = u 2 − 2 v = 1. (B) S^2+C^2=(S+C)^2-2SC=u^2-2v=1. \tag{B} 3. Solve the system From v = 1 4 − u v=\tfrac14-u (by A), substitute in (B): u 2 − 2  ⁣ ( 1 4 − u ) = 1       ⟹       u 2...

  Answer: 92 Answer: 92 Answer: 10 Answer: 75,150,90,150,75 = A,B,C,D,E Answer: 108 Answer: 540 degrees Solution link . Answer: 72 degrees. Answer: 71

Q1. In \(\triangle ABC\), \(a = 5\), \(b = 3\) and \(c = 7\). Then the value of \[ 3\cos C + 7\cos B \] is equal to: Answer: 5 Q2. In \(\triangle ABC\), \(a = 1\), \(b = 2\) and \(\angle C = 60^\circ\). If the area of \(\triangle ABC\) is \(\Delta\), then find \[ 2\Delta\sqrt{3} \] Answer: 3 Q3. In \(\triangle ABC\), if \(b = 20\), \(c = 21\) and \(\sin A = \tfrac{3}{5}\), then \[ a =\;? \] Answer: 13 Q4. If the sides of a triangle are \(5K\), \(6K\), \(5K\) and the in-radius is \(6\), then the value of \(K\) is: Answer: 4 Q5. In \(\triangle ABC\), \(a = 6\), \(b = 8\), \(c = 10\). Then the value of \[ \frac{4R}{r} \] is (where all symbols have their usual meaning): Answer: 10 Q6. In \(\triangle ABC\), \[ a(b^2 + c^2)\cos A \;+\; b(c^2 + a^2)\cos B \;+\; c(a^2 + b^2)\cos C \;=\;\lambda\,abc, \] then the value of \(\lambda\) is: Answer: 3 Q7. In \(\triangle ABC\), with usual notations, if \(8b = 5c\) and \(\angle C = 2\angle B\), then the value of ...

For all the formulae and examples below, 'a' is the side opposite angle A, i.e. BC. Similarly 'b' = AC and 'c' = AB. $$ \cos A = \frac{b^2 + c^2 - a^2}{2\,b\,c} $$ $$ \cos B = \frac{a^2 + c^2 - b^2}{2\,a\,c} $$ $$ \cos C = \frac{a^2 + b^2 - c^2}{2\,a\,b} $$ Example Problems: 1. If $$ a = \sqrt{3},\quad b = \frac12\bigl(\sqrt{6} + \sqrt{2}\bigr),\quad c = \sqrt{2}, $$ then $$ \angle A =\;? $$ Answer: $$ \angle A = 60^\circ$$ Hint: Use Cosine rule 2. If (a+b+c)(a-b+c) = 3ac then $$\angle B = ?$$ Answer: $$60^\circ$$ Hint: Cosine rule 3. If $$ b = 3,\quad c = 4,\quad \angle B = 60^\circ $$ then how many such triangles are possible? Answer: 0 Solution hint: 1. If you use Cosine rule for CosB, you will get a quadratic with 0 solutions. 2. If you use Sine rule, then you will get SinC > 1. 4. $$ \angle B = 30^\circ, c = b\sqrt3, \angle A = ?$$ Answer: $$ \angle A = 90^\circ$$ Hint: Use Sine rule. 5. Prove that $$ \frac{\co...

Q1. In \(\triangle ABC\), \(\angle B = 90^\circ\). If \(AB = 14\) cm and \(AC = 50\) cm then \(\tan A\) equals: Options: (A) \(\tfrac{24}{25}\) (B) \(\tfrac{24}{7}\) (C) \(\tfrac{7}{24}\) (D) \(\tfrac{25}{24}\) Answer: (B) 24/7 Q2. If \(\sin\theta = \tfrac{12}{13}\) then the value of \(\displaystyle \frac{2\cos\theta + 3\tan\theta}{\sin\theta + \tan\theta\,\sin\theta}\) is: Options: (A) \(\tfrac{12}{5}\) (B) \(\tfrac{5}{13}\) (C) \(\tfrac{259}{102}\) (D) \(\tfrac{259}{65}\) Answer: (C) 259/102 Q3. If \(\sec\theta = \dfrac{\sqrt{p^2+q^2}}{q}\) then the value of \(\displaystyle \frac{p\sin\theta - q\cos\theta}{p\sin\theta + q\cos\theta}\) is: Options: (A) \(\tfrac{p}{q}\) (B) \(\tfrac{p^2}{q^2}\) (C) \(\tfrac{p^2 - q^2}{p^2 + q^2}\) (D) \(\tfrac{p^2 + q^2}{p^2 - q^2}\) Answer: (C) \(\tfrac{p^2 - q^2}{p^2 + q^2}\) Q4. The value of \[ 2\tan^2 60^\circ \;-\;4\cos^2 45^\circ\;-\;3\sec^2 30^\circ \] is: Options: (A) 0 (B) 1 (C) 12 (D) 8 Answer: (A) 0 Q5. The value of ...

Compound Angle: $$ \sin(A+B)=\sin A\cos B+\cos A\sin B $$ $$ \sin(A-B)=\sin A\cos B-\cos A\sin B $$ $$ \cos(A+B)=\cos A\cos B-\sin A\sin B $$ $$ \cos(A-B)=\cos A\cos B+\sin A\sin B $$ $$ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} $$ $$ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} $$ Multiple Angle: $$ \sin2\theta = 2\sin\theta\cos\theta $$ $$ \cos2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1 = \cos^2\theta - \sin^2\theta $$ $$ \sin3\theta = 3\sin\theta - 4\sin^3\theta $$ $$ \cos3\theta = 4\cos^3\theta - 3\cos\theta $$ $$ \tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} $$ Range: $$ -1 \le \sin\theta \le 1 \;\;\implies\;\; \sin\theta \in [-1,\,1] $$ $$ -1 \le \cos\theta \le 1 \;\;\implies\;\; \cos\theta \in [-1,\,1] $$ $$ -\infty $$ -\infty $$ |\sec\theta|\ge1 \;\;\implies\;\; \sec\theta \in (-\infty,\,-1] \,\cup\,[1,\,\infty) $$ $$ |\csc\theta|\ge1 \;\;\implies\;\; \csc\theta \in (-\infty,\,-1] \,\cup\,[1,\,\infty) $$ $$ \bigl|a...

Q1. If $$ \sin A - \sqrt{6}\,\cos A = \sqrt{7}\,\cos A $$ then $$ \cos A + \sqrt{6}\,\sin A =\;? $$ Options: (1) \(\sqrt{6}\sin A\)  (2) \(\sqrt{7}\sin A\)  (3) \(\sqrt{6}\cos A\)  (4) \(\sqrt{7}\cos A\) Answer: (2) \(\sqrt{7}\sin A\)  Q2. The minimum value of $$ 7\cos \theta + 24\sin \theta $$ is ? Options: (1) 25  (2) -7  (3) -25  (4) None Answer: -25 Q3. Let $$ \sin(2x)=\frac{1}{7}. $$ Find the numerical value of $$ \sin(x)\sin(x)\sin(x)\sin(x)\;+\;\cos(x)\cos(x)\cos(x)\cos(x). $$ Answer: 97/98 Q4. Let x be a real number such that $$ \sec x - \tan x = 2. $$ Then $$ \sec x + \tan x =\;? $$ Options: (A) 0.1  (B) 0.2  (C) 0.3  (D) 0.4  (E) 0.5 Answer: (E) 0.5 Q5. If $$ \sin(x) = 3\cos(x) $$ then what is $$ \sin(x)\cdot\cos(x)\;? $$ Options: (A) \(\tfrac{1}{6}\)  (B) \(\tfrac{1}{5}\)  (C) \(\tfrac{2}{9}\)  (D) \(\tfrac{1}{4}\)  (E) \(\tfrac{3}{10}\) Answer: (E) 3/10 Q6. In a right triangle the square of the hypot...

Q1. If $$ \sin\theta + \cos\theta = 1 $$ then $$ \sin\theta \cos\theta =\;? $$ Options: (1) 0  (2) 1  (3) 2  (4) \(\frac{1}{2}\) Answer: 0 Q2. Evaluate $$ 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 $$ Options: (1) 2  (2) 0  (3) 4  (4) 6 Answer: 0 Q3. If $$ \csc A + \cot A = \frac{11}{2}, $$ then $$ \tan A =\;? $$ Options: (1) \(\frac{22}{117}\)   (2) \(\frac{2}{11}\)   (3) \(\frac{117}{22}\)   (4) \(\frac{44}{117}\) Answer: \(\frac{44}{117}\)

Q1. Evaluate: $$ 2\bigl(\cos^4 60^\circ + \sin^4 60^\circ\bigr) \;-\;\bigl(\tan^2 60^\circ + \cot^2 45^\circ\bigr) \;+\;3\,\sec^2 30^\circ. $$ Answer: 5/4 Q2. Prove that $$ \sqrt{\;\sec^2\theta \;+\;cosec^2\theta\;} \;=\;\tan\theta\;+\;\cot\theta. $$ Q3. If $$ \frac{\sin^4x}{2}\;+\;\frac{\cos^4x}{3}\;=\;\frac15, $$ then which are the correct options: $$ \begin{aligned} (a)\quad &\tan^2x \;=\;\frac{2}{3},\\ (b)\quad &\frac{\sin^8x}{8}\;+\;\frac{\cos^8x}{27}\;=\;\frac{1}{125},\\ (c)\quad &\tan^2x \;=\;\frac{1}{3},\\ (d)\quad &\frac{\sin^8x}{8}\;+\;\frac{\cos^8x}{27}\;=\;\frac{2}{125}. \end{aligned} $$ Answers: (a), (b) Q4. Prove that $$ \Bigl(1 + \frac{1}{\tan^2\theta}\Bigr)\, \Bigl(1 + \frac{1}{\cot^2\theta}\Bigr) \;=\; \frac{1}{\sin^2\theta \;-\;\sin^4\theta}\,. $$