Correct Answer: $3$

Solution : $\frac{\sin A}{\cot A+\operatorname{cosec} A}-\frac{\sin A}{\cot A-\operatorname{cosec} A}+1$
$=\frac{\sin A}{\frac{\cos A}{\sin A}+\frac{1}{\sin A}}-\frac{\sin A}{\frac{\cos A}{\sin A}-\frac{1}{\sin A}}+1$
$=\frac{\sin^2 A}{\cos A+1}-\frac{\sin^2 A}{\cos A-1}+1$
$=\frac{\sin^2 A}{1+\cos A}+\frac{\sin^2 A}{1-\cos A}+1$
$=\sin^2 A[\frac{1}{1+\cos A}+\frac{1}{1-\cos A}]+1$
$=\sin^2 A[\frac{2}{1-\cos ^2A}]+1$
$=\sin^2 A[\frac{2}{\sin ^2A}]+1$
$=2+1$
$=3$
Hence, the correct answer is $3$.

Correct Answer: Wife

Solution : According to the given information, the family tree is shown below –

Here, the quadrilateral represents the male, and the circular figure represents the female in the figure.

So, according to the above family tree, the lady is the wife of Amit. Hence, the fourth option is correct.

Correct Answer: 14 cm

Solution : Let the second side of the triangle be x cm.
So, first side = 2x cm
Third side = (x + 11) cm
Given: Perimeter of the triangle = 67 cm
So, 2x + x + x + 11 = 67
⇒ 4x = 56
$\therefore$ x = 14 cm
Hence, the correct answer is 14 cm.

Correct Answer: Sarvepalli Radhakrishnan

Solution : The correct option is Sarvepalli Radhakrishnan.

Sarvepalli Radhakrishnan was one of the speakers after Jawaharlal Nehru's historic "Tryst with Destiny," addressed the Parliament on midnight August 15, 1947. Sarvepalli Radhakrishnan stands out as the only one among the options who addressed the Parliament after Nehru on that historic night. He delivered a thoughtful speech emphasizing India's unique cultural heritage and its potential to contribute to the world. This speech added another layer of significance to the momentous occasion.

Correct Answer: QAJ

Solution : Given:
KAP, MIN, IUR, OEL, GOT, ?

The middle term is a vowel with an alternate position (A–I–U–E–O–A) –

So, the required missing term is QAJ. Hence, the second option is correct.

Correct Answer: 730

Solution : Given,
a² + b² = 82
ab = 9
We know,
(a + b)² = a² + b² + 2ab
⇒ (a + b)² = 82 + 2 × 9
⇒ (a + b)² = 100
⇒ (a + b) = 10
Now,
(a + b)³ = a³ + b³ + 3ab(a + b)
⇒ (10)³ = a³ + b³ + 3 × 9 × 10
⇒ 1000 = a³ + b³ + 270
$\therefore$ a³ + b³ = 1000 – 270 = 730
Hence, the correct answer is 730.

Correct Answer: 660

Solution : As per the graph,
The total number of boys in schools A, B, C, D and E = 600 + 450 + 750 + 700 + 800 = 3300
The number of  schools = 5
Average = $\frac{\text{Sum of all the observations}}{{\text{Total number of observations}}}$
The average number of boys in schools = $\frac{3300}{5}$ = 660
Hence, the correct answer is 660.

Correct Answer: 1

Solution : Given: $1 \frac{2}{5}-\left[3 \frac{3}{4} \div\left\{1 \frac{1}{4} \div \frac{1}{2}\left(1 \frac{1}{2} \times 3 \frac{1}{3} \div 1 \frac{1}{3}\right)\right\}\right]$
$=\frac{7}{5}-[\frac{15}{4} \div{\{\frac{5}{4} \div \frac{1}{2}(\frac{3}{2} \times \frac{10}{3} \div \frac{4}{3})}\}]$
$=\frac{7}{5}-[\frac{15}{4} \div{\{\frac{5}{4} \div \frac{1}{2}(\frac{3}{2} \times \frac{10}{3} \times \frac{3}{4})}\}]$
$= \frac{7}{5}-[\frac{15}{4} \div{\{\frac{5}{4} \div \frac{1}{2}\times\frac{15}{4}}\}]$
$= \frac{7}{5}-[\frac{15}{4} \div{\{\frac{5}{2} × \frac{15}{4}}\}]$
$= \frac{7}{5}-[\frac{15}{4} \div\frac{75}{8}]$
$= \frac{7}{5}-\frac{2}{5}$
$= \frac{5}{5}$
$= 1$
Hence, the correct answer is 1.

Correct Answer: 125

Solution : Number of A-type employees in the year 1997 = $\frac{42980 \times 20}{100}$ = 8,596
Number of A-type employees in the year 1998 = $\frac{48640 \times 22}{100}$ = 10,700
∴ Required percentage = $\frac{10700}{8596} \times 100$ ≈ 125%
Hence, the correct answer is 125.

Correct Answer: JMHK

Solution : Given:
RSBG, NPEI, ?, FJKM, BGNO

In the above-given series, subtract 4 and 3 from the place values of the first and second letters, and add 3 and 2 to the place values of the third and fourth letters to obtain the next letter cluster.
RSBG→R – 4 = N; S – 3 = P; B + 3 = E; G + 2 = I
NPEI→N – 4 = J; P – 3 = M; E + 3 = H; I + 2 = K
JMHK→J – 4 = F; M – 3 = J; H + 3 = K; K + 2 = M
FJKM→F – 4 = B; J – 3 = G; K + 3 = N; M + 2 = O

So, JMHK is the missing term of the series. Hence, the third option is correct.