Question

solve this ASAP.....​

Answer 1

Answer:

There is something wrong in the question...I have tried it many times... may be any misprint...plz check it...

Answer 2

CORRECT QUESTION :-

PROVE THAT -

[ tanking ∅ as a ]

$\rm \: {tan \: a - cot \: a = \dfrac{2 \: sin {}^{2}a \: - 1}{sin \: a \times cos \: a} }$

GIVEN :-

$\rm \: {LHS = tan \: a - cot \: a }$

$\rm \: { R HS= \dfrac{2 \: sin {}^{2}a \: - 1}{sin \: a \times cos \: a} }$

TO FIND :-

$\rm \: {tan \: a - cot \: a = \dfrac{2 \: sin {}^{2}a \: - 1}{sin \: a \times cos \: a} }$

SOLUTION :-

now taking LHS :-

we know that

$\implies \boxed{\rm{tan \: x = \dfrac{sin \: x}{cos \: x} }}$

and

$\implies \boxed{\rm{cot\: x = \dfrac{cos \: x}{sin \: x} }}$

hence putting them as follow ,

$\implies \rm{\dfrac{sin \: a}{cos \: a} - \dfrac{cos \: a}{sin \: a} }$

$\implies \rm{ \: \: \dfrac{sin {}^{2} a - {cos}^{2}a \: \: }{cos \: a \: . \: sin \: a} }$

now we know that according to trignometric identies

$\implies \boxed{\rm{ {cos}^{2} \: x = 1 - {sin}^{2}x }}$

hence ,

$\implies \rm{ \: \: \dfrac{sin {}^{2} a - (1 - {sin}^{2}a) \: \: }{cos \: a \: . \: sin \: a} }$

$\implies \rm{ \: \: \dfrac{sin {}^{2} a - 1 + {sin}^{2}a \: \: }{cos \: a \: . \: sin \: a} }$

$\implies \rm{ \dfrac{2sin {}^{2} a - 1 }{sin \: a \: . \:co s\: a} }$

hence , LHS = RHS

and

$\implies \boxed{ \boxed{\rm \: {tan \: a - cot \: a = \dfrac{2 \: sin {}^{2}a \: - 1}{sin \: a \times cos \: a} }}}$

OTHER INFORMATION :-

TRIGNOMETRIC IDENTITIES

• sin²∅ + cos²∅ = 1

• sec²∅ - tan²∅ = 1

• cosec²∅ - cot²∅ = 1

TRIGNOMETRIC RATIOS

• sin ∅ = 1 / cosec ∅

• cos ∅ = 1 / sec ∅

• tan ∅ = 1 / cot ∅

TRIGNOMETRIC COMPLEMENTRY ANGLES

• sin ∅ = cos ( 90 - ∅ )

• cos ∅ = sin ( 90 - ∅ )

• sec ∅ = cosec ( 90 - ∅ )

• cosec ∅ = sec ( 90 - ∅ )

• tan ∅ = cot ( 90 - ∅ )

• cot ∅ = tan ( 90 - ∅ )