Question

tan a + cot a = sec a cosec a, prove..​

Answer 1

$\mathfrak{\underline{\underline{\green{Solution:-}}}}$

To Prove:

tanA + cotA = secA × cosecA

Proof:

Let us consider LHS

LHS:

$\mathsf{ = tanA+cotA }$

As we know

$\boxed{\red{ tanA = \dfrac{sinA}{cosA}}}$

$\boxed{\red{cotA = \dfrac{cosA}{sinA} }}$

$\mathsf{= \dfrac{sinA}{cosA}+ \dfrac{cosA}{sinA}}$

$\mathsf{= \dfrac{{sinA}^{2}+{cosA}^{2}}{sinA \times cosA}}$

By the Identity

$\boxed{\red{ {sinA}^{2}+{cosA}^{2}= 1}}$

$\mathsf{= \dfrac{1}{sinA \times cosA}}$

$\mathsf{= \dfrac{1}{sinA} \times \dfrac{1}{cosA}}$

As we know

$\boxed{\red{ \dfrac{1}{sinA}= cosecA}}$

$\boxed{\red{ \dfrac{1}{cosA}= secA}}$

$\mathsf{= cosecA \times secA}$

$\mathsf{= secA \times cosecA}$

RHS: = $\mathsf{= secA \times cosecA}$

Here

$\boxed{\pink{ LHS = RHS}}$

HENCE PROVED!

Answer 2

ANSWER:

$\sf tan a+cot a=sec a.cosec a$

the above equation is required to prove

let us consider LHS

$\sf tan a+cot a$

$\sf \boxed {tan a = \frac {sina}{cosa}}$

$\sf \boxed {cot a= \frac {cos a}{sina}}$

then

$lhs = \frac{sina}{cosa} + \frac{cosa}{sina}$

when we take the LCM of the two fractions

we get

$= \frac{ {sin}^{2}a + {cos}^{2} a }{cosa \times sina}$

but we know that

$\sf \boxed {sin^2a+cos^2a=1}$

then

$lhs = \frac{1}{cosa \times sina}$

$= \frac{1}{cosa} \times \frac{1}{sina}$

but

$\sf \boxed {\frac {1}{cosa}=seca }$

$\sf \boxed {\frac {1}{sina}=coseca}$

then LHS =seca.coseca

LHS=RHS

hence proved

IDENTITIES USED

$tana = \frac{sina}{cosa}$

$cota = \frac{cosa}{sina}$

${sin}^{2} a + {cos}^{2} a = 1$

$\frac{1}{cosa} = seca$

$\frac{1}{sina} = coseca$