TRIGONOMETRIC IDENTITIES

1. \((\sec A – \tan A)^2 = \dfrac{1 – \sin A}{1 + \sin A}\)
2. \(\dfrac{1 – \cos A}{1 + \cos A} = (\cot A – \csc A)^2\)
3. \(\dfrac{1}{\sec A – 1} + \dfrac{1}{\sec A + 1} = 2 \csc A \cot A\)
4. \(\dfrac{\cos A}{1 – \tan A} + \dfrac{\sin A}{1 – \cot A} = \sin A + \cos A\)
5. \(\dfrac{\csc A}{\csc A – 1} + \dfrac{\csc A}{\csc A + 1} = 2 \sec^2 A\)
6. \(\dfrac{\tan^2 A}{1 + \tan^2 A} + \dfrac{\cot^2 A}{1 + \cot^2 A} = 1\)
7. \(\dfrac{\cot A – \cos A}{\cot A + \cos A} = \dfrac{\csc A – 1}{\csc A + 1}\)
   [NCERT, CBSE 2008]
8. \(\dfrac{1 + \cos \theta – \sin^2 \theta}{\sin \theta (1 + \cos \theta)} = \cot \theta\)
9. (i) \(\dfrac{1 + \cos \theta + \sin \theta}{1 + \cos \theta – \sin \theta} = \dfrac{1 + \sin \theta}{\cos \theta}\)
   (ii) \(\dfrac{\sin \theta – \cos \theta + 1}{\sin \theta + \cos \theta – 1} = \sec \theta – \tan \theta\)
   [CBSE 2001, NCERT]
   (iii) \(\dfrac{\cos \theta – \sin \theta + 1}{\cos \theta + \sin \theta – 1} = \csc \theta + \cot \theta\)
   (iv) \((\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \csc \theta\)
   [NCERT EXEMPLAR]
10. \(\dfrac{1}{\sec A + \tan A} – \dfrac{1}{\cos A} = \dfrac{1}{\cos A} – \dfrac{1}{\sec A – \tan A}\)
    [CBSE 2005]
11. \(\tan^2 A + \cot^2 A = \sec^2 A \csc^2 A – 2\)
12. \(\dfrac{\tan A}{1 + \sec A} – \dfrac{\tan A}{1 – \sec A} = 2 \csc A\)
    [NCERT EXEMPLAR]
13. \(1 + \dfrac{\cot^2 \theta}{1 + \csc \theta} = \csc \theta\)
14. \(\dfrac{\cos \theta}{\csc \theta + 1} + \dfrac{\cos \theta}{\csc \theta – 1} = 2 \tan \theta\)
    [NCERT EXEMPLAR]
15. \((1 + \tan^2 A) + \left( \dfrac{1 + 1}{1 + \tan^2 A} \right) = \dfrac{1}{\sin^2 A – \sin^4 A}\)
    [CBSE 2006C]
16. \(\sin^2 A \cos^2 B – \cos^2 A \sin^2 B = \sin^2 A – \sin^2 B\)
17. (i) \(\dfrac{\cot A + \tan B}{\cot B + \tan A} = \cot A \tan B\)
    (ii) \(\dfrac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B\)
18. \(\cot^2 A \csc^2 B – \cot^2 B \csc^2 A = \cot^2 A – \cot^2 B\)
19. \(\tan^2 A \sec^2 B – \sec^2 A \tan^2 B = \tan^2 A – \tan^2 B\)

LEVEL – 2

20. If \(x = a \sec \theta + b \tan \theta\) and \(y = a \tan \theta + b \sec \theta\), prove that
    \[x^2 – y^2 = a^2 – b^2\]
    [CBSE 2001, 2002C]