Question

trigonometry equations​

Answer 1

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies \frac{1 - {tan}^{2} \theta}{ {cot}^{2} \theta - 1} = {tan}^{2} \theta \\ \\ \red{\underline \bold{To \:Prove:}} \\ \tt: \implies \frac{1 - {tan}^{2} \theta}{ {cot}^{2} \theta - 1} = {tan}^{2} \theta$

• According to given question :

$\tt: \implies \frac{1 - {tan}^{2} \theta}{ {cot}^{2} \theta - 1} = {tan}^{2} \theta \\ \\ \bold{solving \: lhs} \\ \tt: \implies \frac{1 - {tan}^{2} \theta}{ {cot}^{2} \theta - 1} \\ \\ \tt \circ \: {tan}^{2} \theta = {sec}^{2} \theta - 1 \\ \tt: \implies \frac{1 -( {sec}^{2} \theta - 1)}{ {cot}^{2} \theta - 1 } \\ \\ \tt: \implies \frac{1 - {sec}^{2} \theta + 1 }{ {cot}^{2} \theta - 1} \\ \\ \tt \circ \: {cot}^{2} \theta + 1 = {cosec}^{2} \theta \\ \tt: \implies \frac{ - {sec}^{2} \theta }{ {cot}^{2} \theta - ( {cosec}^{2} \theta - {cot}^{2} \theta) } \\ \\ \tt: \implies \frac{ - {sec}^{2} \theta }{ {cot}^{2} \theta - {cosec}^{2} \theta + {cot}^{2} \theta } \\ \\ \tt: \implies \frac{ - {sec}^{2} \theta }{ - {cosec}^{2} \theta } \\ \\ \tt: \implies \frac{1}{ {cos}^{2} \theta } \times {sin}^{2} \theta \\ \\ \tt: \implies \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta } \\ \\ \green{\tt: \implies {tan}^{2} \theta} \\ \\ \green{\tt \: lhs = rhs}\\\\ \red{\huge{\tt Proved}}$

Answer 2

Answer:-

Theta is taken as "A".

(1 - tan² A)/(Cot² A - 1) = tan² A

After cross multiplication we get,

→ (1 - tan² A) = tan² A * Cot² A - (1)(tan² A)

→ (1 - tan² A) = tan² A(1/tan² A) - tan² A

(Since, Cot² A = 1/tan² A)

Cancelling tan² A in RHS we get,

→ 1 - tan² A = 1 - tan² A

→ LHS = RHS.

Hence, Proved.