Question

cot^2(90-A)/tan^2A-1 + cose^2A/sec^2A-cosec^2A = 1/ sin^2-cos^2A​

Answer 1

To prove:

$\dfrac{cot^2(90-A)}{tan^2A-1} +\dfrac{cosec^2A}{sec^2A-cosec^2A} = \dfrac{1}{ sin^2-cos^2A}$

Let us have a look at the few identities:

$1. tan\theta = \dfrac{sin\theta}{cos\theta}$

$2. cot\theta = \dfrac{cos\theta}{sin\theta}$

$3. cosec\theta = \dfrac{1}{sin\theta}$

$4. sec\theta = \dfrac{1}{cos \theta}$

$5. cot(90-\theta)=tan\theta$

$6. sin^2\theta+cos^2\theta=1$

LHS(Left Hand Side):

$\dfrac{cot^2(90-A)}{tan^2A-1} +\dfrac{cosec^2A}{sec^2A-cosec^2A}\\\text{Using } cot(90-\theta) = tan\theta:\\\Rightarrow \dfrac{tan^2A}{tan^2A-1} +\dfrac{cosec^2A}{sec^2A-cosec^2A}$

Let us put the values of $tanA, cotA, cosecA$ and $sec A$ from the above formulas (1) to (4):

$\Rightarrow \dfrac{\dfrac{sin^2A}{cos^2A}}{\dfrac{sin^2A}{cos^2A}-1} +\dfrac{\dfrac{1}{sin^2A}}{\dfrac{1}{cos^2A}-\dfrac{1}{sin^2A}}\\\Rightarrow \dfrac{\dfrac{sin^2A}{cos^2A}}{\dfrac{sin^2A-cos^2A}{cos^2A}} +\dfrac{\dfrac{1}{sin^2A}}{\dfrac{sin^2A-cos^2A}{cos^2Asin^2A}}\\\Rightarrow \dfrac{sin^2A\times cos^2A}{cos^2A\times (sin^2A-cos^2A)}+\dfrac{cos^2A\times sin^2A}{sin^2A\times (sin^2A-cos^2A)}$

$\Rightarrow \dfrac{sin^2A}{(sin^2A-cos^2A)}+\dfrac{cos^2A}{(sin^2A-cos^2A)}\\\Rightarrow \dfrac{sin^2A+cos^2A}{(sin^2A-cos^2A)}\\\Rightarrow \dfrac{1}{(sin^2A-cos^2A)} (\because sin^2A+cos^2A =1)$

Hence, proved that:

$\dfrac{cot^2(90-A)}{tan^2A-1} +\dfrac{cosec^2A}{sec^2A-cosec^2A} = \dfrac{1}{ sin^2-cos^2A}$