Question

a
cos²0 sina
(1 – tan o) (sin 0 - cos 0)
=(1 + sin
cos )​

Answer 1

Question :-

Prove that

$\frac{ { \cos}^{2} \theta }{(1 - \tan \theta)} + \frac{ { \sin}^{3} \theta}{( \sin \theta - \cos \theta)} = (1 + \sin \theta \cos \theta )$

Io

Proof :-

→ We will try to show that left hand side is equal to right hand side .

Formula used :-

$\star \: \sf{ {a}^{3} - {b}^{ 3} = (a - b) ( {a}^{2} + {b}^{2} + ab)} \: \\ \\ \star \: \: { \sin}^{2} \theta + { \cos}^{2} \theta \: = 1 \\ \\ \star \: \: \tan \theta \: = \frac{ \sin \theta}{ \cos \theta \: }$

Explanation :-

Taking left hand side ,

$\rightarrow\frac{ { \cos}^{2} \theta }{(1 - \tan \theta)} + \frac{ { \sin}^{3} \theta}{( \sin \theta - \cos \theta)} \\ \\ \rightarrow \: \frac{ { \cos}^{2} \theta }{(1 - \frac{ \sin \theta}{ \cos \theta} )} + \frac{ { \sin}^{3} \theta}{( \sin \theta - \cos \theta)} \: \\ \\ \rightarrow \: \frac{ { \cos}^{2} \theta }{( \frac{ \cos \theta - \sin \theta}{ \cos \theta} )} + \frac{ { \sin}^{3} \theta}{( \sin \theta - \cos \theta)} \: \\ \: \\ \rightarrow \: \frac{ { \cos}^{3} \theta }{( \cos \theta - \sin \theta)} + \frac{ {\sin}^{3} \theta}{( \sin \theta - \cos \theta)} \: \\ \\ \text{ we \: can \: write \: it} \\ \\ \rightarrow \: \frac{ { \cos}^{3} \theta }{( \cos \theta - \sin \theta)} - \frac{ {\sin}^{3} \theta}{( \cos \theta - \sin \theta)} \: \: \\ \\ \rightarrow \: \frac{ { \cos}^{3} \theta - { \sin}^{3} \theta}{( \cos \theta \: - \sin \theta)} \\ \\ \because \sf{ {a}^{3} - {b}^{ 3} = (a - b)( {a}^{2} + {b}^{2} + ab)} \\ \therefore \\ \\ \rightarrow \: \frac{( \cos \theta - \sin \theta)( { \cos}^{2} \theta + { \sin}^{2} \theta + \sin \theta \cos \theta }{( \cos \theta \: - \sin \theta \: )} \\ \\ \because \: { \sin}^{2} \theta + { \cos}^{2} \theta \: = 1 \\ \therefore \\ \rightarrow \: 1 + \sin \theta \: \cos \theta \:$

Left hand side = right hand side

(LHS = RHS )

hence proved : )

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