Correct Answer: $2-\sqrt{3}$

Solution : Given that $\cos \left(2 \theta+54^{\circ}\right)=\sin \theta$ and $0^{\circ}<\left(2 \theta+54^{\circ}\right)<90^{\circ}$
$⇒\cos \left(2 \theta+54^{\circ}\right)=\cos(90^{\circ}-\theta)$
Solving for $\theta$,
$⇒ \left(2 \theta+54^{\circ}\right)= (90^{\circ}-\theta)$
$⇒3 \theta=36^{\circ}$
$⇒\theta=12^{\circ}$
$⇒\tan 5 \theta=\tan \left(60^{\circ}\right)=\sqrt{3}$
$⇒\operatorname{cosec} \frac{5 \theta}{2}=\operatorname{cosec} \left(30^{\circ}\right)=2$
Now, $\frac{1}{\tan 5 \theta+\operatorname{cosec}\frac{5 \theta}{2}}$
$=\frac{1}{\sqrt{3}+2}$
$=\frac{1}{\sqrt{3}+2}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=2-\sqrt{3}$
Hence, the correct answer is $2-\sqrt{3}$.

Correct Answer: much

Solution : The third option is the correct answer: much.

"much" is an adjective or adverb used to indicate a large quantity, degree, or extent of something. "Very" expresses a great degree, "Many" is an adjective indicating a large countable quantity and

Correct Answer: Upliftment of untouchables or Dalits

Solution : The correct answer is the Upliftment of untouchables or Dalits.

The term Sarvodaya was coined by Mahatma Gandhi to signify the upliftment of all sections of society. He was highly influenced by John Ruskin's book Unto This Last and translated it.

Correct Answer: awareness

Solution : The correct choice is the third option.

Awareness means knowledge or perception about something, in this case, about oneself.

This helps in conveying the intended meaning of the sentence, which is that human life is a unique blessing that is dependent on how aware you

Correct Answer: Brahmo Samaj

Solution : The correct option is Brahmo Samaj.

Raja Ram Mohan Roy founded the Brahmo Sabha, which was later known as the Brahmo Samaj. Raja Ram Mohan Roy established the Brahmo Samaj to promote monotheism, the worship of the formless divine, and opposing idol worship.

Correct Answer: $96 \pi$

Solution : The radius of the base of a cylindrical tank, $r$ = 4 m
Let the height of the cylinder be $h$.
According to the question,
Three times the sum of the areas of its two circular faces is twice the area of its curved

Correct Answer: Carnatic Music

Solution : The correct option is Carnatic Music

Kuchipudi, a classical dance form that originated in the state of Andhra Pradesh in India, is traditionally accompanied by Carnatic music. Carnatic music is a form of classical music that has its roots in South India. In this

Correct Answer: $\frac{\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$

Solution : $
\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)
$
Multiplying and dividing by $(\mathrm{k}+\frac{1}{\mathrm{k}})$
= $
\frac{(\mathrm{k}+\frac{1}{\mathrm{k}})\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $
\frac{(\mathrm{k}^2-\frac{1}{\mathrm{k}^2})\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $
\frac{(\mathrm{k}^4-\frac{1}{\mathrm{k}^4})\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})} 
$
= $
\frac{(\mathrm{k}^8-\frac{1}{\mathrm{k}^8})\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $
\frac{(\mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}})\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $\frac{(\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}})\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $\frac{(\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}})}{(\mathrm{k}+\frac{1}{\mathrm{k}})} $
Hence, the correct answer is $\frac{(\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}})}{(\mathrm{k}+\frac{1}{\mathrm{k}})}$.

Correct Answer: –1

Solution : Given:
$x^{2}+y^{2}+2x+1=0$
⇒ $x^{2}+2x+1+y^{2}=0$
⇒ $(x+1)^2+y^2=0$
Since the addition of the squares of the two terms is zero,
So, both the terms are individually zero.
Therefore, $(x+1)^2=0$ and $y^2=0$.
⇒ $x=-1$ and $y=0$
Now, $x^{31}+y^{35}$
= $(-1)^{31}+(0)^{35}$
= –1
Hence, the correct answer is –1.

Correct Answer: (3)

Solution : The error lies in the third part of the sentence.

The preposition that should follow "senior" is to as adjectives ending in "-ior", such as junior, senior, inferior, and superior, are followed by to and not than.

Therefore, the correct sentence is: "She does