Question

find the answer please
prove that: ​

Answer 1

Answer:

cos²x/1-sinx/cosx

cos³x/cosx-sinx

sin³x/sinx-cosx

(cos³x/cosx-sinx)-(sin³x/cosx-sinx)

(cos³x-sin³x)/(cosx-sinx)

(cosx-sinx)(cos²x+sinx.cosx+sin²x)/(cosx-sinx)

sin²x+cos²x+sinx.cosx

1+sinx.cosx= R.H.S= proved..

Answer 2

Here, we've been asked to prove the trigonometric Identity:

$\qquad\qquad\bigstar\quad\tt{\dfrac{{cos}^{2}\theta}{(1-tan\theta)}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}=1+sin\theta\times cos\theta}$

From here we get,

$\bullet\quad\bold{LHS}=\blue{\sf{\dfrac{{cos}^{2}\theta}{(1-tan\theta)}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}}}$

$\bullet\quad\bold{RHS}=\pink{\sf{1+sin\theta\times cos\theta}}$

Let's start working on LHS first. $\\\quad\longrightarrow\quad\tt{LHS=\blue{\dfrac{{cos}^{2}\theta}{(1-tan\theta)}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}}}$

We know, that the reciprocal of cos θ times sin θ equals to tan θ. That is, $\\\qquad\qquad\leadsto\tt{tan\theta=\dfrac{sin\theta}{cos\theta}}$

On applying this, $\\\; \; \longrightarrow\quad\tt{LHS=\dfrac{{cos}^{2}\theta}{\left(1-\dfrac{sin\theta}{cos\theta}\right)}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}}$

Multiplying cos θ and 1. $\\\; \; \longrightarrow\quad\tt{LHS=\dfrac{{cos}^{2}\theta}{\left(\dfrac{cos\theta-sin\theta}{cos\theta}\right)}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}}$ $\\\; \; \longrightarrow\quad\tt{LHS=\dfrac{{cos}^{2}\theta\times cos\theta}{cos\theta-sin\theta}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}}$

Performing multiplication in numerator.$\\\; \; \longrightarrow\quad\tt{LHS=\dfrac{{cos}^{3}\theta}{(cos\theta-sin\theta)}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}}$ $\\\; \; \longrightarrow\quad\tt{LHS=\dfrac{{cos}^{3}\theta-{sin}^{3}\theta}{(cos\theta-sin\theta)}}$

We know, $\\\qquad\qquad\leadsto\tt{{a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})}$

On applying the identity in numerator.$\\\; \; \longrightarrow\quad\tt{LHS=\dfrac{(cos\theta-sin\theta)({cos}^{2}\theta+cos\theta.sin\theta+{sin}^{2}\theta}{(cos\theta-sin\theta)}}$

Performing division. $\\\; \; \longrightarrow\quad\tt{LHS=({sin}^{2}\theta+{cos}^{2}\theta+cos\theta.sin\theta)}$

We know, sum of sin² θ and cos² θ equals to 1. That is, $\\\qquad\qquad\leadsto\tt{{sin}^{2}\theta+{cos}^{2}\theta=1}$

On applying the identity.

$\; \; \longrightarrow\quad\tt{LHS=\pink{1+sin\theta.cos\theta}}$

Which is equal to RHS. Hence,

$\; \; \longrightarrow\quad\sf{\blue{LHS}=\pink{RHS}}$

That is, $\\\twoheadrightarrow\quad\underline{\boxed{\sf{\dfrac{{cos}^{2}\theta}{(1-tan\theta)}+\dfrac{{sin}^{3}\theta}{(sin\theta-cos\theta)}=1+sin\theta\times cos\theta}}}$

Hence, proved!