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**Q.1. Find the value of****Ans.**

We know that

[âˆµ tan^{- 1}(- x) = - tan ^{- 1} x]

Hence,**Q.2. Evaluate****Ans.**

Hence,**Q.3. Prove that****Ans.**

L.H.S.

= cot [cot^{â€“ 1} (7)] = 7 R.H.S.

Hence proved.**Q.4. Find the value of****Ans.**

Hence, **Q.5. Find the value of****Ans.**

We know that

Hence,

Q.6. Show that **Ans.**

L.H.S. 2 tan^{â€“ 1} (â€“ 3) = â€“ 2 tan^{â€“ 1} (3)

Hence proved.**Q.7. Find the real solutions of the equation****Ans.**

â‡’ x^{2} + x = 0 â‡’ x(x + 1) = 0

â‡’ x = 0 or x + 1 = 0 â‡’ x = 0 or x = â€“ 1

Hence the real solutions are x = 0 and x = â€“ 1.**Alternate Method:**

â‡’ x^{2} + x + 1 = 1 â‡’ x^{2} + x = 0

â‡’ x(x + 1) = 0 â‡’ x = 0 or x + 1 = 0

âˆ´ x = 0, x = â€“ 1**Q.8. Find the value of the expression ****Ans.**

Hence,**Q.9. If 2 tan ^{â€“1} (cos Î¸) = tan^{â€“1} (2 cosec Î¸), then show that Î¸ =**

2 tan

â‡’ cos Î¸ sin Î¸ = sin

â‡’ cos Î¸ sin Î¸ â€“ sin

â‡’ sin Î¸ = 0 or cos Î¸ - sin Î¸ = 0 â‡’ sin Î¸ = 0 or 1 - tan Î¸ = 0

â‡’ Î¸ = 0 or tan Î¸ = 1 â‡’ Î¸ = 0Â° or

Hence proved.

L.H.S.

R.H.S.

L.H.S. = R.H.S. Hence proved.

**Q.11. Solve the following equation****Ans.**

Given that

â‡’

Squaring both sides we get,

â‡’

Hence,

Long Answer (L.A.)**Q.12. Prove that****Ans.**

L.H.S.

Put x^{2} = cos Î¸ âˆ´ Î¸ = cos^{â€“ 1} x^{2}

[Dividing the Nr. and Den. by cos Î¸/2]

Hence proved.

Q.13. Find the simplified form of**Ans.**

Given that

Put

= cos^{â€“ 1} [cos (y â€“ x)] = y â€“ x**Q.14. Prove that****Ans.**

L.H.S

Using sin^{â€“ 1} x + sin^{â€“ 1} y =

R.H.S. Hence proved.

Q.15. Show that**Ans.**

Now

â‡’

â‡’

Hence proved.**Q.16. Prove that****Ans.**

âˆ´

Hence,**Q.17. Find the value of****Ans.****Q.18. Show thatand justify why the other value****is ignored?****Ans.**

To prove that

[âˆ´ tan (tan^{â€“ 1} Î¸) = Î¸]**Q.19. If a _{1}, a_{2}, a_{3},...,a_{n} is an arithmetic progression with common difference d, then evaluate the following expression.**

If a

âˆ´ d = a

â‡’ tan [tan

â‡’ tan [tan

Objective Type Questions

Principal value branch of cos

Principal value branch of cosec

as cosec

Hence, the correct answer is (d).

Given that 3 tan

â‡’ 2 tan

âˆ´ x = 1

Hence, the correct answer is (b).

Q.23. The value of

Hence, the correct answer is (d).

(a) [0, 1]

(b) [â€“1, 1]

(d) [0, Ï€]

The given function is cos

Let f(x) = cos

â€“ 1 â‰¤ 2x â€“ 1 â‰¤ 1 â‡’ - 1 + 1 â‰¤ 2x â‰¤ 1 + 1

0 â‰¤ 2x â‰¤ 2 â‡’ 0 â‰¤ x â‰¤ 1

âˆ´ domain of the given function is [0, 1].

Hence, the correct answer is (a)

(b) [â€“1, 1]

(d) none of these

Let

â‡’ 0 â‰¤ x - 1 â‰¤ 1 â‡’ 1 â‰¤ x â‰¤ 2 â‡’ x âˆˆ [1, 2]

Hence, the correct answer is (a).

(d) 1

Given that

â‡’

â‡’

â‡’

â‡’

Hence, the correct answer is (b).

Given that sin [2 tan

= sin [sin

= 0.96

Hence, the correct answer is (c).

Hence, the correct answer is (a).

Hence, the correct answer is (b).

Given that tan

Hence, the correct answer is (a).**Q.31. If****where a, x âˆˆ ]0, 1,then the value of x is**

(b) a/2

â‡’ 4 tan

â‡’

Hence, the correct answer is (d).

We have,

Let

âˆ´

Hence, the correct answer is (d).

We have,

Let

â‡’

Hence, the correct answer is (b).

Here, we have 2 tan

Hence, the correct answer is (a).

Solution.

We have cos

â‡’ cos

â‡’ cos

â‡’ Î± = cos Ï€, Î² = cos Ï€ and Î³ = cos Ï€

âˆ´ Î± = â€“ 1, Î² = â€“ 1 and Î³ = â€“ 1

Which gives a = Î² = Î³ = â€“1

So Î± (Î² + Î³) + Î²( Î³+ Î±) + Î³(Î± + Î²)

â‡’ (â€“ 1)(â€“ 1 â€“ 1) + (â€“ 1)(â€“ 1 â€“ 1) + (â€“ 1)(â€“ 1 â€“ 1)

â‡’ (â€“ 1)(â€“ 2) + (â€“ 1)(â€“ 2) + (â€“ 1)(â€“ 2) â‡’ 2 + 2 + 2 â‡’ 6

Hence, the correct answer is (c).

Solution.

Which does not satisfy for any value of x.

Hence, the correct answer is (d).

Solution.

Here, given that cos

â‡’ sin [cos

We know that â€“ 1 â‰¤ x â‰¤ 1

Hence, the correct answer is (c).

Hence, Principal value of

Hence, the value of

Given that

Hence, the value of x is âˆš3 .

Letâ‡’sec x =

Since, the domain of sec

Hence, sec

Q.42. The principal value of tan

Hence the principal value of tan

Hence, the value of cos

Q.44. The value of cos (sin

Hence, the value of cos (sin

Hence, the value of the given expression is 1.

â‡’ y = 2 tan

â‡’ y = 4 tan

â‡’ â€“ 2Ï€ < y < 2Ï€

Hence, the value of y is (â€“ 2Ï€, 2Ï€).

The given result is true when xy > â€“ 1.

cot

False.

We know that all inverse trigonometric functions are restricted over their domains.

False.

We know that cos

So (cos

True.

We know that all trigonometric functions are restricted over their domains to obtain their inverse functions.

True.

True.

We know that the domain and range are interchanged in the graph of inverse trigonometric functions to that of their corresponding trigonometric functions.

False.

Given that

â‡’

â‡’ n > p â‡’ n > 3.14

Hence, the value of n is 4.

Q.55. The principal value of sin

True.

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