Question

prove that (1+tanø+secø) (1+cotø-cosecø) =2​

Answer 1

Solution :-

We have to prove that,

(1+tanΦ+secΦ)(1+cotΦ-cosec Φ) = 2

Solving LHS,

(1+sinΦ/cosΦ+1/cosΦ)(1+cosΦ/sinΦ-1/sinΦ)

(cosΦ+sinΦ+1/cosΦ)(sinΦ+cosΦ-1/sinΦ)

(cosΦ+sinΦ+1)(sinΦ+cosΦ-1)/cosΦ.sinΦ

= (cosΦ + sinΦ)² - 1 / cosΦ * sinΦ

= cos²Φ+sin²Φ+ 2cosΦ*sinΦ - 1 /cosΦ.sinΦ

= 1 + 2cosΦsinΦ - 1 / cosΦsinΦ

= 2cosΦsinΦ/cosΦsinΦ

= 2

Hence, Proved $.$

Reciprocal Relation and trigonometric identities :-

• sinΦ = 1/cose Φ , Cosec Φ = 1/sinΦ
• cosΦ = 1/secΦ , sec Φ = 1/cos Φ
• tanΦ = 1/cotΦ , cot Φ = 1/tan Φ
• tanΦ = sinΦ/cosΦ
• cot Φ = cosΦ/sinΦ

Identities :-

• Sin²Φ + cos²Φ = 1

• 1 + cot²Φ = cosec²Φ

• 1 + tan²Φ = sec²Φ

Answer 2

Solution -

We have,

➝ (1 + tanθ + sinθ) (1 + cotθ - cosecθ) = 2

⠀

Taking L.H.S

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( 1 + \dfrac{sin \theta}{cos \theta} + \dfrac{1}{cos \theta} \bigg)\: \bigg( 1 + \dfrac{cos \theta}{sin \theta} - \dfrac{1}{sin \theta} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: { \bigg( \dfrac{cos \theta + sin \theta + 1}{cos \theta} \bigg) \: \bigg( \dfrac{sin \theta + cos \theta - 1}{sin \theta} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{(cos \theta + sin \theta + 1) (sin \theta + cos \theta - 1)}{cos \theta sin \theta} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{(cos \theta + sin \theta)^2 - 1}{cos \theta sin \theta} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{cos^2 \theta + sin^2 \theta + 2 cos \theta sin \theta - 1}{cos \theta sin \theta} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{1 + 2cos \theta sin \theta - 1}{cos \theta sin \theta} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{2cos \theta sin \theta }{cos \theta sin \theta} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{2 \cancel{ cos \theta sin \theta}}{\cancel{cos \theta sin \theta}} \bigg)}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {2}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {R.H.S}$

⠀

➝ L.H.S = R.H.S

HENCE PROVED

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