Question

please say me the identities in trigonometry

Answer 1

cos2A + sin2A : 1
1 + tan2A : sec2A
cot2A + 1 : cosec2A

Answer 2

⭐Basic and Pythagorean Identities

\csc(x) = \dfrac{1}{\sin(x)}csc(x)=sin(x)1​

\sin(x) = \dfrac{1}{\csc(x)}sin(x)=csc(x)1​

\sec(x) = \dfrac{1}{\cos(x)}sec(x)=cos(x)1​

\cos(x) = \dfrac{1}{\sec(x)}cos(x)=sec(x)1​

\cot(x) = \dfrac{1}{\tan(x)} = \dfrac{\cos(x)}{\sin(x)}cot(x)=tan(x)1​=sin(x)cos(x)​

\tan(x) = \dfrac{1}{\cot(x)} = \dfrac{\sin(x)}{\cos(x)}tan(x)=cot(x)1​=cos(x)sin(x)​

⭐Angle-Sum and -Difference Identities

sin(α + β) = sin(α) cos(β) + cos(α) sin(β)

sin(α – β) = sin(α) cos(β) – cos(α) sin(β)

cos(α + β) = cos(α) cos(β) – sin(α) sin(β)

cos(α – β) = cos(α) cos(β) + sin(α) sin(β)

\tan(\alpha + \beta) = \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}tan(α+β)=1−tan(α)tan(β)tan(α)+tan(β)​

\tan(\alpha - \beta) = \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \tan(\beta)}tan(α−β)=1+tan(α)tan(β)tan(α)−tan(β)​


⭐Double-Angle Identities

sin(2x) = 2 sin(x) cos(x)

cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1

\tan(2x) = \dfrac{2 \tan(x)}{1 - \tan^2(x)}tan(2x)=1−tan2(x)2tan(x)​

⭐Half-Angle Identities

\sin\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(x)}{2}}sin(2x​)=±21−cos(x)​​

\cos\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 + \cos(x)}{2}}cos(2x​)=±21+cos(x)​​

\tan\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(x)}{1 + \cos(x)}}tan(2x​)=±1+cos(x)1−cos(x)​​

= \dfrac{1 - \cos(x)}{\sin(x)}=sin(x)1−cos(x)​

= \dfrac{\sin(x)}{1 + \cos(x)}=1+cos(x)sin(x)​
⭐Sum Identities

\sin(x) + \sin(y) = 2 \sin\left(\dfrac{x + y}{2}\right) \cos\left(\dfrac{x - y}{2}\right)sin(x)+sin(y)=2sin(2x+y​)cos(2x−y​)

\sin(x) - \sin(y) = 2 \cos\left(\dfrac{x + y}{2}\right) \sin\left(\dfrac{x - y}{2}\right)sin(x)−sin(y)=2cos(2x+y​)sin(2x−y​)

\cos(x) + \cos(y) = 2 \cos\left(\dfrac{x + y}{2}\right) \cos\left(\dfrac{x - y}{2}\right)cos(x)+cos(y)=2cos(2x+y​)cos(2x−y​)

\cos(x) - \cos(y) = -2 \sin\left(\dfrac{x + y}{2}\right) \sin\left(\dfrac{x - y}{2}\right)cos(x)−cos(y)=−2sin(2x+y​)sin(2x−y​)

⭐Product Identities

\sin(x) \cos(y) = \frac{1}{2} \big[\sin(x + y) + \sin(x - y)\big]sin(x)cos(y)=21​[sin(x+y)+sin(x−y)]

\cos(x) \sin(y) = \frac{1}{2} \big[\sin(x + y) - \sin(x - y)\big]cos(x)sin(y)=21​[sin(x+y)−sin(x−y)]

\cos(x) \cos(y) = \frac{1}{2} \big[\cos(x - y) + \cos(x + y)\big]cos(x)cos(y)=21​[cos(x−y)+cos(x+y)]

\sin(x) \sin(y) = \frac{1}{2} \big[\cos(x - y) - \cos(x + y)\big]sin(x)sin(y)=21​[cos(x−y)−cos(x+y)]


⭐✔HOPE IT HELPS YOU⭐✔