MCQ 11 Mark
If in $\text{a}\triangle\text{ABC},\tan\text{B}+\tan\text{C}=6,$ then $\cot\text{A}\cot\text{B}\cot\text{C}=$
AnswerCorrect option: C. $\frac16$
In $\triangle\text{ABC},$
$\text{A+B+C}=\pi$
We know that $\tan(\text{A+B+C)}=\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}}{1-\tan\text{A}\tan\text{B}-\tan\text{B}\tan\text{C}-\tan\text{C}\tan\text{A}}$ and $\tan\pi=0.$
$\therefore\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}=0$
$\tan\text{A}+\tan\text{B}+\tan\text{C}=\tan\text{A}\tan\text{B}\tan\text{C}$
If $\tan\text{A}+\tan\text{B}+\tan\text{C}=6,\tan\text{A}\tan\text{B}\tan\text{C}=6$
$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac16$
$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}=\frac16$
View full question & answer→MCQ 21 Mark
If $\tan\theta_1\tan\theta_2=\text{k},$ then $\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}=$
- ✓
$\frac{1+\text{k}}{1-\text{k}}$
- B
$\frac{1-\text{k}}{1+\text{k}}$
- C
$\frac{\text{k}+1}{\text{k}-1}$
- D
$\frac{\text{k}-1}{\text{k}+1}$
AnswerCorrect option: A. $\frac{1+\text{k}}{1-\text{k}}$
$\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}$
$=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-0\sin\theta_1\sin\theta_2}$
Dividing numerator and denominator by $\cos\theta_1\cos\theta_2,$ we get:
$\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}$
$=\frac{1+\text{k}}{1-\text{k}}$
View full question & answer→MCQ 31 Mark
If $\text{A}-\text{B}=\frac\pi4,$ then $(1+\tan\text{A})(1-\tan\text{B})$ is equal to:
Answer$\tan(\text{A - B})=\tan\frac\pi4$
$\Rightarrow\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}=1$
$\Rightarrow\tan\text{A}-\tan\text{B}=1+\tan\text{A}\tan\text{B}\cdots(1)$
Now,
$(1+\tan\text{A})(1-\tan\text{B})=1+\tan\text{A}-\tan\text{B}-\tan\text{A}\tan\text{B}$
$=1+1+\tan\text{A}\tan\text{B}-\tan\text{A}\tan\text{B}$
$=2$
View full question & answer→MCQ 41 Mark
The value of $\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\sin^2\Big(\frac\pi6-\text{x}\Big)$ is:
AnswerCorrect option: A. $\frac{1}{2}\cos2\text{x}$
$\cos^2\Big(\frac\pi6+\text{x}\Big)-\sin^2\Big(\frac{\pi}{6}-\text{x}\Big)$
$=\cos\Big(\frac{\pi}{6}+\text{x}+\frac{\pi}{6})-\text{x}\Big)\cos\Big(\frac{\pi}{6}+\text{x}-\frac{\pi}{6}+\text{x}\Big)$ $\Big[\text{Using}\cos(\text{A+B})\cos(\text{A-B})=\cos^2\text{A}-\sin^2\text{B}\Big]$
$=\cos\frac{2\pi}{6}\cos2\text{x}$
$=\frac12\cos2\text{x}$ $\Big[\text{As}\cos\frac\pi3=\frac12\Big]$
View full question & answer→MCQ 51 Mark
If $\tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ=2\text{k},$ then k =
- A
$-1$
- B
$\frac12$
- ✓
$\frac{-1}{2}$
- D
AnswerCorrect option: C. $\frac{-1}{2}$
$\tan135^\circ=\tan(90^\circ+45^\circ)$
$=-\tan45^\circ$
$=-1$
Or, $\tan(69^\circ+66^\circ)=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}$
$\Rightarrow-1=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}$
$\Rightarrow\tan69^\circ+\tan66^\circ-\tan69^\circ+\tan66^\circ=-1$
$\therefore2\text{k}=-1$
$\Rightarrow\text{k}=\frac{-1}{2}$
View full question & answer→MCQ 61 Mark
If $\cot(\alpha+\beta)=0,$ then $\sin(\alpha+2\beta)$ is equal to:
- ✓
$\sin\alpha$
- B
$\cos2\beta$
- C
$\cos\alpha$
- D
$\sin2\alpha$
AnswerCorrect option: A. $\sin\alpha$
Given:
$\cot(\alpha+\beta)=0$
$\Rightarrow\frac{\cos(\alpha+\beta)}{\sin(\alpha+\beta)}=0$
$\Rightarrow\cos(\alpha+\beta)=0$
$\Rightarrow\alpha+\beta=\frac\pi2$
$\therefore\sin(\alpha+2\beta)=\sin(\alpha+\alpha+\beta)$
$=\sin\alpha$
View full question & answer→MCQ 71 Mark
If $\text{A+B+C}=\pi,$ then $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$ is equal to:
Answer$\pi=180^\circ$
Using $\tan(180^\circ-\text{A})=-\tan\text{A},$ we get:
$\text{C}=\pi-(\text{A+B})$
Now, $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan[\pi-\text{(A+B)}]}{\tan\text{A}\tan\text{B}\tan[\pi-\text{(A+B)}]}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan\text{(A+B)}}{-\tan\text{A}\tan\text{B}\tan\text{(A+B)}}$
$=\frac{\tan\text{A}+\tan\text{B}-\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}{-\tan\text{A}\tan\text{B}\times\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}-\tan\text{A}-\tan\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=\frac{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=1$
View full question & answer→MCQ 81 Mark
If $\tan(\text{A}-\text{B})=1,\sec(\text{A+B})=\frac{2}{\sqrt{3}},$ then the smallest positive value of B is:
- A
$\frac{25\pi}{24}$
- ✓
$\frac{19\pi}{24}$
- C
$\frac{13\pi}{24}$
- D
$\frac{11\pi}{24}$
AnswerCorrect option: B. $\frac{19\pi}{24}$
Given:
$\tan(\text{A - B})=1$ and $\sec(\text{A+B})=\frac{2}{\sqrt{3}}$
$\Rightarrow\text{A - B}=\frac{\pi}{4}\cdots(1)$ and $\text{A + B}=\frac{\pi}{4}\cdots(2)$
Adding these equations we get:
$2\text{A}=\frac{\pi}{4}+\frac\pi6$
$\Rightarrow\text{A}=\frac{5\pi}{24}$
$\Rightarrow$ Smallest possible value of $\text{B}=\pi-\frac{5\pi}{24}=\frac{19\pi}{24}.$
View full question & answer→MCQ 91 Mark
The value of $\sin^2\frac{5\pi}{12}-\sin^2\frac{\pi}{12}$ is:
- A
$\frac12$
- ✓
$\frac{\sqrt{3}}{2}$
- C
$1$
- D
$0$
AnswerCorrect option: B. $\frac{\sqrt{3}}{2}$
$\frac{5\pi}{12}=75^\circ,\frac{\pi}{12}=15^\circ$
$\sin^275^\circ-\sin^215^\circ$
$=\sin^275^\circ-\cos^275^\circ$ $[\sin(90^\circ-\theta)=\cos\theta]$.
Now, $\sin75^\circ=\sin(45^\circ+30^\circ)$
$=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\cos75^\circ=\cos(45^\circ+30^\circ)$
$=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Hence,
$\sin^275^\circ-\cos^275^\circ=\Big(\frac{\sqrt{3}+1}{2\sqrt{2}}\Big)^2-\Big(\frac{\sqrt{3}-1}{2\sqrt{2}}\Big)^2$
$=\frac{3+1+2\sqrt{3}-3-1+2\sqrt{3}}{8}$
$=\frac{4\sqrt{3}}{8}$
$=\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 101 Mark
If $\tan20^\circ+\tan40^\circ+\sqrt{3}\tan20\circ\tan40^\circ$ is equal to:
- A
$\frac{\sqrt{3}}{4}$
- B
$\frac{\sqrt{3}}{2}$
- ✓
$\sqrt{3}$
- D
$1$
AnswerCorrect option: C. $\sqrt{3}$
$\tan20^\circ+\tan40^\circ+\sqrt{3}\tan20^\circ\tan40^\circ$
$=\tan60^\circ(1-\tan20^\circ\tan40^\circ)+\tan60^\circ\tan20^\circ\tan40^\circ$
$=\tan60^\circ-\tan60^\circ\tan20^\circ\tan40^\circ+\tan60^\circ\tan20^\circ\tan40^\circ$
$=\tan60^\circ$
$=\sqrt{3}$
View full question & answer→MCQ 111 Mark
The maximum value of $\sin^2\Big(\frac{2\pi}{3}+\text{x}+\sin^2\Big(\frac{2\pi}{3}-\text{x}\Big)$ is:
- A
$\frac12$
- ✓
$\frac32$
- C
$\frac14$
- D
$\frac34$
AnswerCorrect option: B. $\frac32$
$\frac{2\pi}{3}=120^\circ$
Let $\text{f(x)}=\sin^2(90+30+\text{x})+\sin^2(90+30-\text{x})$
$=[\cos(30+\text{x})]^2+[\cos(30=\text{x})]^2$ $[\text{Using }\sin(90+\text{A})=\cos\text{A}]$
$=\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2+\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2$
$=\frac{\sqrt{3}}{2}\cos^2\text{x}-\frac14\sin^2\text{x}-\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}+\frac34\cos^2\text{x}+\frac14\sin^2\text{x}+\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}$
$=\frac{3}{2}\cos^2\text{x}-\frac12\sin^2\text{x}$
$=\frac{3}{2}(1-\sin^2\text{x})+\frac12\sin^2\text{x}$
$=\frac32-\frac{3}{2}\sin^2\text{x}+\frac12\sin^2\text{x}$
$=\frac32-\sin^2\text{x}$.
For f(x) to be maximum, $\sin^2\text{x}$ must have minimum value, which is 0.
$\therefore\frac32$ is the maximum value of f(x).
View full question & answer→MCQ 121 Mark
If $\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x}),$ then $\sin2\text{x}=$
- A
$\pm\frac34$
- B
$\pm\frac43$
- ✓
$\pm\frac13$
- D
AnswerCorrect option: C. $\pm\frac13$
$\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x})$
As we know that $\sin\text{x}=-\cos\Big(\frac\pi2+\text{x}\Big)$
$\Rightarrow-\cos\Big(\frac\pi2+\pi\cos\text{x}\Big)=\cos(\pi\sin\text{x})$
$\Rightarrow\frac{-\pi}{2}-\pi\cos\text{x}=\pi\sin\text{x}$
$\Rightarrow\pi\sin\text{x}-\pi\cos\text{x}=\frac12$
$\Rightarrow\sin\text{x}-\cos\text{x}=\frac12$
Squaring both sides we get,
$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\frac14$
$\Rightarrow1+\sin2\text{x}=\frac{1}{4}$
$\Rightarrow\sin2\text{x}=\frac13$
$\therefore\sin2\text{x}=\pm\frac13$
View full question & answer→MCQ 131 Mark
If $\tan\alpha=\frac{\text{x}}{\text{x}+1}$ and $\tan\beta=\frac{1}{2\text{x}+1},$ then $\alpha+\beta$ is equal to:
- A
$\frac\pi2$
- B
$\frac\pi3$
- C
$\frac\pi6$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
It is given that $\tan\alpha=\frac{\text{x}}{\text{x}+1}$ and $\tan\beta=\frac{1}{2\text{x}+1}.$
$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$=\frac{\frac{\text{x}}{\text{x}+1}+\frac{1}{2\text{x+1}}}{1-\frac{\text{x}}{\text{x}+1}\times\frac{1}{2\text{x+1}}}$
$=\frac{\frac{\text{x}(2\text{x}+1)+(\text{x}+1)}{(\text{x}+1)(2\text{x}+1)}}{\frac{(\text{x}+1)(2\text{x}+1)-\text{x}}{(\text{x}+1)(2\text{x}+1)}}$
$=\frac{2\text{x}^2+\text{x}+\text{x}+1}{2\text{x}^2+3\text{x}+1+\text{x}}$
$=\frac{2\text{x}^2+2\text{x}+1}{2\text{x}^2+2\text{x}+1}$
$=1$
$\therefore\alpha+\beta=\frac\pi4\ (\tan\frac\pi4=1)$
Hence, the correct answer is option D.
View full question & answer→MCQ 141 Mark
If $\cos\text{P}=\frac{1}{7}$ then $\cos\text{Q}=\frac{13}{14},$ where P and Q both are acute angles. Then, the value of P - Q is:
- A
$\frac{\pi}{6}$
- ✓
$\frac\pi3$
- C
$\frac\pi4$
- D
$\frac{5\pi}{12}$
AnswerCorrect option: B. $\frac\pi3$
$\cos\text{P}=\frac17,\cos\text{Q}=\frac{13}{14}$
$\therefore\sin\text{P}=\sqrt{1-\frac{1}{49}}=\frac{4\sqrt{3}}{7}$ and $\sin\text{Q}=\sqrt{1-\frac{169}{196}}=\frac{3\sqrt{3}}{14}$
Hence, $\tan\text{P}=4\sqrt{3},\tan\text{Q}=\frac{3\sqrt{3}}{13}$
$\cos(\text{P - Q})=\cos\text{P}\cos\text{Q}+\sin\text{P}\sin\text{Q}$
$=\frac{1}{7}\times\frac{13}{14}+\frac{4\sqrt{3}}{7}\times\frac{3\sqrt{3}}{14}$
$=\frac{13+36}{98}$
$=\frac{49}{98}$
$\therefore\cos(\text{P - Q})=\frac12$
$\Rightarrow\text{P - Q}=\cos^{-1}\frac12$
$\Rightarrow\text{P - Q}=60^\circ$
Hence, the correct answer is option B.
View full question & answer→MCQ 151 Mark
$\tan3\text{A}-\tan2\text{A}-\tan\text{A}$ is equal to:
- ✓
$\tan3\text{A}\tan2\text{A}\tan\text{A}$
- B
$-\tan3\text{A}-\tan2\text{A}\tan\text{A}$
- C
$\tan\text{A}\tan2\text{A}\tan2\text{A}\tan3\text{A}-\tan3\text{A}\tan\text{A}$
- D
AnswerCorrect option: A. $\tan3\text{A}\tan2\text{A}\tan\text{A}$
$3\text{A}=2\text{A}+\text{A}$
$\Rightarrow\tan3\text{A}=\tan(2\text{A}+\text{A})$
$\Rightarrow\tan3\text{A}=\tan(2\text{A}+\text{A})=\frac{\tan2\text{A}+\tan\text{A}}{1-\tan2\text{A}\tan\text{A}}$
$\Rightarrow\tan3\text{A}-\tan3\text{A}\tan2\text{A}\tan\text{A}=\tan2\text{A}+\tan\text{A}$
$\Rightarrow\tan3\text{A}-\tan2\text{A}-\tan\text{A}=\tan3\text{A}\tan2\text{A}\tan\text{A}$
View full question & answer→MCQ 161 Mark
If $\tan\text{A}=\frac{\text{a}}{\text{a}+1}$ and $\text{B}=\frac{1}{2\text{a}+1},$ then the value of A + B is:
- A
$0$
- B
$\frac{\pi}{2}$
- C
$\frac\pi3$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
$\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$
$=\frac{\frac{\text{a}}{\text{a}+1}+\frac{1}{2\text{a}+1}}{1-\frac{\text{a}}{(\text{a}+1)(2\text{a}+1)}}$
$=\frac{2\text{a}^2+\text{a}+\text{a}+1}{2\text{a}^2+3\text{a}+1-\text{a}}$
$=\frac{2\text{a}^2+2\text{a}+1}{2\text{a}^2+2\text{a}+1}$
$=1$
$\therefore \text{ A+B}=\tan^{-1}(1)=\frac\pi4.$
View full question & answer→MCQ 171 Mark
If $\cos(\text{A}-\text{B})=\frac35$ and $\tan\text{A}\tan\text{B}=2,$ then
- ✓
$\cos\text{A}\cos\text{B}=\frac15$
- B
$\cos\text{A}\cos\text{B}=-\frac15$
- C
$\sin\text{A}\sin\text{B}=-\frac15$
- D
$\sin\text{A}\sin\text{B}=-\frac15$
AnswerCorrect option: A. $\cos\text{A}\cos\text{B}=\frac15$
$\tan\text{A}\tan\text{B}=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}=2\ (\text{Given})\cdots(1)$
Also,
$\cos(\text{A - B})=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac35+\sin\text{A}\sin\text{A}=\frac35$
$\therefore\sin\text{A}\sin\text{B}=\frac35-\cos\text{A}\cos\text{B}\cdots(2)$
Substituting eq (2) in eq (1), we get:
$\Rightarrow\frac{\frac35-\cos\text{A}\cos\text{B}}{\cos\text{A}\cos\text{B}}=2$
$\Rightarrow3\cos\text{A}\cos\text{B}=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac15$
View full question & answer→MCQ 181 Mark
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$ is equal to:
- ✓
$\tan55^\circ$
- B
$\cot55^\circ$
- C
$-\tan35^\circ$
- D
$-\cot35^\circ$
AnswerCorrect option: A. $\tan55^\circ$
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$
$=\frac{1+\tan10^\circ}{1-\tan10^\circ}$ $[$Dividing the numerator and denominator by $\cos10^\circ]$
$=\frac{\tan45^\circ+\tan10^\circ}{1-\tan45^\circ\times\tan10^\circ}$
$=\tan(45^\circ+10^\circ)$ $\Big[$Using $\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$=\tan55^\circ$
View full question & answer→MCQ 191 Mark
If $3\sin\text{x}+4\cos\text{x}=5,$ then $4\sin\text{x}-3\cos\text{x}=$
Answer$3\sin\text{x}+4\cos\text{x}=5$
$\frac{3}{5}\sin\text{x}+\frac{4}{5}\cos\text{x}=1$
Let $\cos\alpha=\frac35$ and $\sin\alpha=\frac45.$
$\therefore\cos\alpha\sin\text{x}+\sin\alpha\cos\text{x}=1$
$\Rightarrow\sin(\alpha+\text{x})=\sin\frac\pi2$
$\Rightarrow\alpha+\text{x}=\frac\pi2$
$\Rightarrow\text{x}=\frac{\pi}{2}-\alpha\cdots(1)$
We have to find the value of $4\sin\text{x}-3\cos\text{x}.$
$4\sin\Big(\frac\pi2-\alpha\Big)-3\cos\Big(\frac\pi3-\alpha\Big)$ ...{From eq (1)}
$=4\cos\alpha-3\sin\alpha$
$=4\times\frac35-3\times\frac45$ $\Big(\because\cos\alpha=\frac35$ and $\sin\alpha=\frac{4}{5}\Big)$
$=0$
View full question & answer→MCQ 201 Mark
If $\tan\theta=\frac12$ and $\tan\phi=\frac13,$ then the value of $\theta+\phi$ is:
- A
$\frac\pi6$
- B
$\pi$
- C
$0$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
It is given that $\tan\theta=\frac12$ and $\tan\phi=\frac13.$
Now,
$\tan(\theta+\phi)=\frac{\tan+\tan\phi}{1-\tan\theta\tan\phi}$
$=\frac{\frac12+\frac13}{1-\frac12\times\frac13}$
$=\frac{\frac{5}{6}}{\frac56}$
$=1$
$\therefore\theta+\phi=\frac\pi4$ $\Big(\tan\frac\pi4=1\Big)$
Hence, the correct answer is option D.
View full question & answer→MCQ 211 Mark
The value of $\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ+\text{A})\cos(54^\circ-\text{A})$ is:
- A
$\sin2\text{A}$
- ✓
$\cos2\text{A}$
- C
$\cos3\text{A}$
- D
$\sin3\text{A}$
AnswerCorrect option: B. $\cos2\text{A}$
$\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ+\text{A})\cos(54^\circ-\text{A})$
$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin[90^\circ-(54^\circ+\text{A})]\sin[90^\circ-(54^\circ-\text{A})]$
$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin(36^\circ-\text{A})\sin(36^\circ-\text{A})$
$=\cos(36^\circ-\text{A}-36^\circ+\text{A})$$\big[\text{Using}\cos(\text{A - B) formula} \big]$
$=\cos2\text{A}$
View full question & answer→MCQ 221 Mark
If $\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a},$ then $\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)=$
- A
$a^2+1$
- B
$a^2+2$
- ✓
$a^2-2$
- D
AnswerCorrect option: C. $a^2-2$
- $a^2-2$
Solution:
Given:
$\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}$
$\Rightarrow\Big[\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)\Big]^2=\text{a}^2$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)+2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}^2$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{\tan45^\circ-\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\times\frac{\tan45^\circ+\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\Big]$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{1^\circ-\tan\text{x}}{1+\tan\text{x}}\times\frac{1^\circ+\tan\text{x}}{1-\tan\text{x}}\Big]$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big(\frac{1-\tan^2\text{x}}{1-\tan^2\text{x}}\Big)$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2$ View full question & answer→MCQ 231 Mark
If $\text{A+B+C}=\pi,$ then $\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})$ is equal to:
Answer$\pi=180^\circ$
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\pi-(\text{A+B}))-\sin\text{B}\sin(\pi(\text{A+B}))}{\cos\text{A}}$
We know that, $\cos(\pi-\theta)=-\cos\theta$ and $\sin(\pi-\theta)=\sin\theta,$
$\therefore\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\text{A+B})-\sin\text{B}\sin(\text{A+B})}{\cos\text{A}}$
Now, using the identities $\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$ and $\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B},$ we get
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}\cos\text{B}^2+\cos\text{B}\sin\text{A}\sin\text{B}-\sin\text{B}\sin\text{A}\cos\text{B}-\sin^2\cos\text{A}}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}(\cos^2\text{B}+\sin^2\text{B})}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}}{\cos\text{A}}=-1$
View full question & answer→