Question

tan theta / 1- cot theta + cot theta / 1- tan theta = 1+sec theta× cosec theta​

Answer 1

Answer:

Step-by-step explanation:

Answer

1−cotθtanθ​+1−tanθcotθ​

=1−tanθ1​tanθ​+1−tanθtanθ1​​

=tanθ−1tan2θ​−tanθ−1tanθ1​​

=tanθ−1tan2θ−tanθ1​​=tanθ(tanθ−1)tan3θ−1​

=tanθ(tanθ−1)(tanθ−1)(tan2θ+tanθ+1)​

=tanθtan2θ+tanθ+1​

=tanθ+1+cotθ

=cosθsinθ​+sinθcosθ​+1

=1+cosθ⋅sinθsin2θ+cos2θ​

=1+cosθsinθ1​

=1+secθ⋅cosecθ.

solution

Answer 2

Answer:

Step-by-step explanation:

Given:

$\tt \dfrac{tan\: \theta}{1-cot\: \theta} +\dfrac{cot\: \theta}{1-tan\: \theta} =1+sec\: \theta\times cosec\: \theta$

To Prove:

LHS = RHS

Proof:

Taking the LHS of the equation,

$\tt \dfrac{tan\: \theta}{1-cot\: \theta} +\dfrac{cot\: \theta}{1-tan\: \theta}$

We know,

cot θ = 1/tan θ

Substituting the identities we get,

$\tt \implies \dfrac{tan\: \theta}{1-\dfrac{1}{tan\: \theta} } +\dfrac{\dfrac{1}{tan\: \theta} }{1-tan\: \theta}$

Cross multiplying we get,

$\tt \implies \dfrac{tan\: \theta}{\dfrac{tan\: \theta-1}{tan\: \theta} } +\dfrac{\dfrac{1}{tan\: \theta} }{1-tan\: \theta}$

$\tt \implies \dfrac{tan\: \theta(tan\: \theta)}{1-tan\: \theta} + \dfrac{\dfrac{1}{tan\: \theta} }{-(tan\: \theta-1)}$

$\tt \implies \dfrac{tan^{2}\: \theta }{tan\: \theta-1} - \dfrac{\dfrac{1}{tan\: \theta} }{tan\: \theta-1}$

Adding the fractions since the denominators are same,

$\tt \implies \dfrac{tan^{2}\: \theta-\dfrac{1}{tan\: \theta} }{tan\: \theta-1}$

Cross multiplying,

$\tt \implies \dfrac{tan^{3}\: \theta-1 }{tan\: \theta} \times \dfrac{1}{tan\: \theta-1}$

We know,

(a³ - b³) = (a - b) (a² + ab + b²)

Apply the identity,

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

Cancelling tan θ - 1 on both numerator and denominator,

$\tt \implies \dfrac{tan^{2} \: \theta + tan\: \theta + 1}{tan\: \theta}$

$\tt \implies \dfrac{tan^{2}\: \theta }{tan\: \theta}+\dfrac{tan\: \theta}{tan\: \theta} +\dfrac{1}{tan\: \theta}$

We know 1/tan θ = cot θ

Hence,

$\tt \implies tan\: \theta + 1 + cot\: \theta$

Converting tan and cot in terms of cos and sin

$\implies \tt \dfrac{sin\: \theta}{cos\: \theta} + \dfrac{cos\: \theta}{sin\: \theta} + 1$

Cross multiply,

$\tt \implies \dfrac{sin^{2}\: \theta + cos^{2}\: \theta }{sin\: \theta\times cos\: \theta} + 1$

We know

Sin²θ + cos ²θ = 1

$\tt \implies \dfrac{1 }{sin\: \theta\times cos\: \theta} + 1$

$\tt \implies \dfrac{1 }{sin\: \theta} \times \dfrac{1}{cos\: \theta} + 1$

We know that,

1/sin θ = cosec θ

1/cos θ = sec θ

Hence,

$\tt \implies 1 + sec\: \theta \times cosec\: \theta$

= RHS

Hence proved.