Question

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Proof the given identity.

(cosecx - sinx)(secx - cosx) = 1/(tanx + cotx)

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Answer 1

Step-by-step explanation:

We will use simplification in both the sides, for easy process underway.

So, as hinted, simplifying the RHS will help.

We have, RHS as :-

$\sf{\dfrac{1}{tanx+cotx}}$

We can also write tanx and cotx in terms of sinx and cosx.

$\sf{\dfrac{1}{\dfrac{sinx}{cosx}+ \dfrac{cosx}{sinx}}}$

$\sf{\dfrac{1}{\dfrac{sin^2 x + cos^2 x}{cosx . sinx}}}$

sin²x + cos²x = 1 , applying it:-

$\sf{\dfrac{1}{\dfrac{1}{cosx . sinx}}}$

Resultant RHS - $\sf{cosx . sinx }$

Moving on to LHS,

$\sf{(\dfrac{1}{sinx} - sinx)(\dfrac{1}{cosx} - cosx)}$

Converted cosecx into 1/sinx and secx into 1/cosx

$\sf{(\dfrac{1 - sin^2 x}{sinx})(\dfrac{1 - cos^2 x}{cosx})}$

$\sf{\dfrac{(1 - sin^2 x)(1 - cos^2 x)}{sinx . cosx}}$

Breaking the brackets,

$\sf{\dfrac{1 - cos^2 x - sin^2 x + sin^2x . cos^2 x }{sinx . cosx}}$

Applying sin²x + cos²x = 1 ,

$\sf{\dfrac{1 - 1 sin^2x . cos^2 x }{sinx . cosx}}$

$\sf{\dfrac{(sinx . cosx)(sinx . cosx) }{sinx . cosx}}$

Cutting one of the brackets off with denominator,

We get cosx . sinx

cosx . sinx = cosx . sinx

LHS = RHS

Hence, proved.

Answer 2

Question:-

• Proof the given identity.$\sf (cosecx - sinx)(secx - cosx) = \dfrac { 1 } { (tanx + cotx) }$

To Find:-

• Prove that.

Solution:-

We have RHS as:-

• $\sf \: \dfrac { 1 } { tanx + cotx }$

Now ,

$\longrightarrow\sf \: \dfrac { 1 } { \dfrac { sinx } { cosx } + \dfrac { cosx } { sinx } }$

Here ,

• $\sf \: { sin }^{ 2 } x + { cos }^{ 2 } x = 1$

$\longrightarrow\sf \: \dfrac { 1 } { \dfrac { 1 } { cosx.sinx } }$

Now ,

$\longrightarrow\sf \: LHS = ( \dfrac { 1 } { sinx } - sinx )( \dfrac { 1 } { cosx } - cosx )$

$\longrightarrow\sf \: \dfrac { 1 - { cos }^{ 2 }x - { sin }^{ 2 }x + { sin }^{ 2 }x \times { cos }^{ 2 }x }{ sinx \times cosx }$

$\longrightarrow\sf \: \dfrac { ( sinx.cosx )( sinx.cosx ) } { ( sinx.cosx ) }$

$\longrightarrow\sf \: sinx.cosx$

$\longrightarrow\sf \: LHS = RHS$

• Hence proved