**NCERT Exercise 7.4: Questions 2, 3, 4, 5, and 6 Solutions**

**Question 2:**
Find the value of `tan 75° - cot 15°`.

**Solution:**
To find the value of `tan 75° - cot 15°`, we need to evaluate the individual trigonometric ratios first.

1. `tan 75°`:
Using the formula `tan A = sin A / cos A`, we can find the value of `tan 75°` as follows:
`tan 75° = sin 75° / cos 75°`.
Now, we need to express sin and cos in terms of complementary angles.
`sin 75° = cos (90° - 75°) = cos 15°`
`cos 75° = sin (90° - 75°) = sin 15°`
Therefore, `tan 75° = cos 15° / sin 15°`.

2. `cot 15°`:
Using the formula `cot A = cos A / sin A`, we can find the value of `cot 15°` as follows:
`cot 15° = cos 15° / sin 15°`.

Now, we substitute the values of `tan 75°` and `cot 15°` back into the original expression:
`tan 75° - cot 15° = (cos 15° / sin 15°) - (cos 15° / sin 15°) = 0`.

Hence, the value of `tan 75° - cot 15°` is 0.

**Question 3:**
If `tan A = √3`, find the value of `sin A + cos A`.

**Solution:**
Given, `tan A = √3`.
Using the Pythagorean identity, `tan² A + 1 = sec² A`, we can find the value of `sec A`.
Plugging in the given value of `tan A`:
`(√3)² + 1 = sec² A`
`3 + 1 = sec² A`
`4 = sec² A`
Taking the square root of both sides, we get:
`2 = sec A`.

Now, we can find the values of `sin A` and `cos A` using the identities:
`sin A = 1 / sec A = 1 / 2`
`cos A = √(1 - sin² A) = √(1 - 1/4) = √(3/4) = √3/2`.

Finally, we can find the value of `sin A + cos A`:
`sin A + cos A = 1/2 + √3/2 = (1 + √3)/2`.

Hence, the value of `sin A + cos A` is `(1 + √3)/2`.

**Question 4:**
If `cot θ = √2`, find the value of `tan θ`.

**Solution:**
Given, `cot θ = √2`.
Using the identity `cot θ = 1 / tan θ`, we can find the value of `tan θ`.
Plugging in the given value of `cot

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