Question

tan A+cot A=sec A cosec A

Answer 1

To Prove :-

tan A + cot A = sec A cosec A

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Proof :-

Let us Consider LHS

$:\implies\sf\; tan \; A + cot\; A$

As We know that,

$\hookrightarrow\sf\; tan\; A \; = \; \large\frac{sin\;A}{cos\;A}$

$\hookrightarrow\sf \; cot \; A \; = \; \large\frac{cos\; A}{sin\; A}$

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$:\implies\sf\large\frac{sin\; A}{cos\;A} \; = \; {cos \; A}\;{sin\; A}$

Taking LCM

$:\implies\sf\LARGE\frac{sin^2\;A \; + \;cos^2\;A}{cos \; A sin\; A}$

By using the identity

$\large\boxed{\sf{sin^2 \; A\; +\; cos^2\; A \;=\;1}}}$

$:\implies\sf\Large\frac{1}{sin\; A \; cos\; A}\; =\; \frac{1}{cos\;A}\times\frac{1}{sin\;A}$

$:\implies\sf\; sec\; A \; cosec\; A$

LHS = RHS

$:\implies\sf\;tan \;A + cot \;A = \sec \;A \;cosec\; A$

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Answer 2

Correct question= Is it possible to prove (tan A + cot A) (cosec A - sin A) (sec A - cos A) = 1?

Solution⬇️

LHS =(tanA+cotA)(cosecA-SinA)(secA-CosA)

tanA+cotA=(SinA/CosA)+(CosA/SinA)

= (sin²Α+cos²Α)/cosAsinA

= 1/cosAsinA

CosecA-SinA=(1/SinA)-SinA

= (1-sin²Α)/SinA= cos²Α/sinA

SecA-cosA=(1/cosA)-cosA=(1-cos²Α)/cos A

=sin²Α/cosA

(tanA+cotA)(cosecA-SinA)(secA-cosA)

=(1/sinAcosA)(cos²Α/SinA)(sin²Α/CosA)

=1=RHS