Question

Prove that (1 + cot teta - cosec teta )(1 + tan teta + sec teta) = 2.​

Answer 1

Answer:

2. Ans.

Step-by-step explanation:

->LHS= [1+cos/sin-1/sin][1+1/cos+sin/cos]

= [cos+sin-1][cos+sin+1]/cos*sin

=[cos+sin]^2-(1)^2/cos*sin

=1+2cos•sin-1/cos•sin

now 1 and -1 will be cut down

=2cos•sin/cos•sin

cos•sin will cut down

= 2 ....Ans.

HOPE THIS WILL HELP YOU!

Answer 2

To Prove :

• (1 + cot theta - cosec theta )(1 + tan theta + sec theta) = 2.

Solution :

LHS :(1 + cotØ - cosecØ)(1 + tanØ + secØ)

we know that,

• cotØ = cosØ/sinØ
• tanØ = sinØ/cosØ

$\rm \: = (1 + \frac{cos \: \theta}{sin \: \theta} - cosec \: \theta)(1 + \frac{sin \: \theta}{cos \: \theta }+ sec \: \theta)$

• cosecØ = 1/sinØ
• secØ = 1/cosØ

$\rm \: = (1 + \frac{cos \: \theta}{sin \: \theta} - \frac{1}{sin \: \theta} )(1 + \frac{sin \: \theta}{cos \: \theta} + \frac{1}{cos \: \theta})$

$\rm \: = ( \frac{sin \: \theta + cos \: \theta - 1}{sin \: \theta} )( \frac{cos \: \theta + sin \: \theta + 1}{cos \: \theta} )$

$\rm \: = \frac{ \{(sin \: \theta + cos \: \theta) - 1 \}. \{(sin \: \theta + cos \: \theta) + 1 \}}{sin \: \theta.cos \: \theta}$

$\rm \: = \frac{{(sin \: \theta + cos \: \theta)}^{2} - 1}{ sin \: \theta.cos \: \theta}$

$\rm \: = \frac{{sin {}^{2} \: \theta + cos {}^{2} \: \theta + 2sin \: \theta \: cos \: \theta} - 1}{ sin \: \theta.cos \: \theta}$

we know that,

• sin²Ø + cos²Ø = 1

$\rm \: = \frac{1 + 2sin \theta \: cos \theta - 1}{sin \theta \: cos \theta}$

$\rm \: = \frac{2 \cancel{sin \theta \: cos \theta}}{ \cancel{sin \theta \: cos \theta}}$

= 2

Hence, Proved

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