Question

Prove that
$( \sin(tetta) + \sec(tetta) ) {}^{2}$
+
$( \cos(tetta) + \csc(tetta)) {}^{2}$
=
$(1 + \sec(tetta) \csc(tetta) ) {}^{2}$
​

Answer 1

To prove

(sin∅ + sec∅)² + (cos∅ + cosec∅)² = (1 + sec∅cosec∅)²

Proof

Taking LHS,

Write sec∅ as 1/cos∅ and cosec∅ as 1/sin∅

→ (sin∅ + 1/cos∅)² + (cos∅ + 1/sin∅)²

→ sin²∅ + 1/cos²∅ + 2sin∅/cos∅ + cos²∅ + 1/sin²∅ + 2cos∅/sin∅

(using, (a + b)² = a² + b² + 2ab)

→ (sin²∅ + cos²∅) + (1/cos²∅ + 1/sin²∅) + 2(sin∅/cos∅ + cos∅/sin∅)

using sin²∅ + cos²∅ = 1,

→ 1 + (cos²∅ + sin²∅)/sin²∅cos²∅ + 2(sin²∅ + cos²∅)/sin∅cos∅

→ 1 + 1/sin²∅cos²∅ + 2/sin∅cos∅

→ 1 + sec²∅cosec²∅ + 2sec∅cosec∅

→ (1 + sec∅cosec∅)²

[Using, a² + b² + 2ab = (a + b)²]

Hence proved.

Answer 2

$\huge{\underline{\underline{\bf{Solution:-}}}}$

${\bullet{\textsf{Let's take LHS:-}}}$

$\implies{ \sf{(sin\theta + sec\theta) {}^{2} + (cos \theta + cosec \theta) {}^{2} }} \\ \\ \implies{ \sf{(sin\theta + \frac{1}{cos\theta}) {}^{2} + (cos \theta + \frac{1}{sin \theta} ) {}^{2} }} \\ \\{ \sf{\bullet{ using \: identity:- }}}\\ { \sf{ sin {}^{2} \theta + cos {}^{2} \theta = 1 }} \\ \\ \implies{ \sf{sin {}^{2} \theta + \frac{1}{cos {}^{2} \theta} + 2 \frac{sin \theta}{cos \theta} + cos {}^{2} \theta + \frac{1}{sin {}^{2} \theta } + 2 \frac{cos \theta}{sin \theta} }} \\ \\ \implies{ \sf{(sin {}^{2} \theta + cos {}^{2} \theta) + \frac{1}{cos {}^{2} \theta} + \frac{1}{sin {}^{2} \theta } + 2( \frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} )}} \\ \\ \implies{ \sf{1 + \frac{cos {}^{2} \theta + sin {}^{2} \theta }{sin {}^{2} \theta + cos {}^{2} \theta } + 2 \frac{(sin {}^{2} \theta + cos {}^{2} \theta) }{sin \theta \: cos \theta} }} \\ \\ \implies{ \sf{1 + \frac{1}{sin {}^{2} \theta \: cos {}^{2} \theta } + \frac{2}{sin \theta \: cos \theta} }} \\ \\ \implies { \sf{1 + sec {}^{2} \theta \: cos {}^{2} \theta + 2sec \theta \: cosec \theta}} \\ \\ {\bullet { \sf{using \: identity:-}}}\\ \: { \tt{ (a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab}} \\ \\ \implies{ \bf{ (1 + sec \theta \: cosec \theta) {}^{2} }}$

$\huge{\textsf{\# Hence proved.....}}$