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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Statement1 : ...

Question

Statement1 : $c o s^{2} A$ + $s i n^{2} A$ = 1

Statement2 : $s e c^{2} A$ + $t a n^{2} A$ = 1

Statement1 is true and Statement2 is false

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Both Statement1 are Statement2 are true

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Both Statement1 are Statement2 are false

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Statement2 is true and Statement1 is false

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Solution

The correct option is A
Statement1 is true and Statement2 is false

Identity : $s i n^{2} A$ + $c o s^{2} A$ = 1

Identity : $s e c^{2} A$ - $t a n^{2} A$ = 1

Suggest Corrections

Similar questions

If A = $30^{\circ},$ verify that:

(i) $s i n 2 A = \frac{2 t a n A}{1 + t a n^{2} A}$ (ii) $c o s 2 A = \frac{1 - t a n^{2} A}{1 + t a n^{2} A}$ (iii) $t a n 2 A = \frac{2 t a n A}{1 - t a n^{2} A}$

\frac{\sqrt{1 - c o s^{2} A}}{\sqrt{s e c^{2} A - 1}}

\times

\frac{\sqrt{1 + t a n^{2} A}}{\sqrt{1 - s i n^{2} A}} =

Q. Prove that:

\frac{{cos}^{2} A + {tan}^{2} A - 1}{{sin}^{2} A} = {tan}^{2} A

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $s i n^{2} A - c o s^{2} A$

3.tan A = $\frac{2 t a n \frac{A}{2}}{1 - t a n^{2} \frac{A}{2}}$

4. $s i n^{2} \frac{A}{2}$ = 1 - cosA

5. $t a n^{2} A$ = $\frac{1 - c o s 2 A}{1 + c o s 2 A}$

6. $s i n^{2} A$ = $\frac{2 t a n A}{1 - t a n^{2} A}$

Q. If A =

30^{\circ}

, then

cos 2 A = \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A} = {cos}^{2} A - {sin}^{2} A

State true or false.

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Statement1 : cos2A + sin2A = 1 Statement2 : sec2A + tan2A = 1

The correct option is A Statement1 is true and Statement2 is false Identity : sin2A + cos2A = 1 Identity : sec2A - tan2A = 1

Statement1 : $c o s^{2} A$ + $s i n^{2} A$ = 1

Statement2 : $s e c^{2} A$ + $t a n^{2} A$ = 1

The correct option is A
Statement1 is true and Statement2 is false

Identity : $s i n^{2} A$ + $c o s^{2} A$ = 1

Identity : $s e c^{2} A$ - $t a n^{2} A$ = 1