Question

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Answer 1

$\large\underline{\sf{Given \:Question - }}$

In ∆ PQR, right angled at Q, in which PQ = 4 units, PR = 5 units. What is the value of tan²R - cot²P ?

$\large\underline{\sf{Solution-}}$

Given that,

A ∆ PQR, right angled at Q, such that PQ = 4 units and PR = 5 units.

Now, In ∆ PQR

By Pythagoras theorem, we have

$\rm :\longmapsto\: {PR}^{2} = {PQ}^{2} + {QR}^{2}$

On substituting the values, we get

$\rm :\longmapsto\: {5}^{2} = {4}^{2} + {QR}^{2}$

$\rm :\longmapsto\: 25 = 16 + {QR}^{2}$

$\rm :\longmapsto\: 25 - 16 = {QR}^{2}$

$\rm :\longmapsto\: 9= {QR}^{2}$

$\bf\implies \:QR = 3 \: units$

Now,

$\rm :\longmapsto\:tanR = \dfrac{opposite \: side}{adjacent \: side}$

$\rm :\longmapsto\:tanR = \dfrac{PQ}{QR}$

$\rm :\longmapsto\:tanR = \dfrac{4}{3}$

Now,

$\rm :\longmapsto\:cotP = \dfrac{adjacent \: side}{opposite \: side}$

$\rm :\longmapsto\:cotP = \dfrac{PQ}{QR}$

$\rm :\longmapsto\:cotP = \dfrac{4}{3}$

So, Consider

$\rm :\longmapsto\: {tan}^{2}R \: - \: {cot}^{2}P$

$\rm \: = \: {\bigg[\dfrac{4}{3} \bigg]}^{2} - {\bigg[\dfrac{4}{3} \bigg]}^{2}$

$\rm \: = \: \dfrac{16}{9} - \dfrac{16}{9}$

$\rm \: = \: 0$

So,

$\rm :\longmapsto\: \red{ \boxed{ \sf{ \: \: \: \:{tan}^{2}R \: - \: {cot}^{2}P = 0 \: \: \: \: }}}$

Alternative Method :-

Given that,

A ∆ PQR, right angled at Q

$\bf\implies \:\angle \:P + \angle \:Q + \angle \:R = 180 \degree$

$\rm :\longmapsto\: \:\angle \:P + 90\degree+ \angle \:R = 180 \degree$

$\rm :\longmapsto\: \:\angle \:P + \angle \:R = 180 \degree - 90\degree$

$\rm :\longmapsto\: \:\angle \:P + \angle \:R = 90 \degree$

$\rm :\longmapsto\: \: \angle \:R = 90 \degree - \angle \:P$

$\rm :\longmapsto\: \: tanR =tan( 90 \degree - P)$

$\rm :\longmapsto\: \: tanR =cot P$

On squaring both sides, we get

$\rm :\longmapsto\: \: tan {}^{2} R =cot {}^{2}P$

$\rm :\longmapsto\: \: tan {}^{2}R - cot {}^{2}P = 0$