Question

Prove that (1+tan theta+sec theta)(1+cot theta -cosec theta)=2

Answer 1

Step-by-step explanation:

Given: (1+tan∅+sec∅)(1+cot∅-cosec∅)

To prove: =2

then,

$→(1+tan∅+sec∅)(1+cot∅-cosec∅)$

$→(1+\frac{sin∅}{cos∅}+\frac{1}{cos∅})(1+\frac{cos∅}{sin∅}-\frac{1}{sin∅})$

$(\frac{cos∅+sin∅+1}{cos∅})(\frac{sin∅+cos∅-1}{sin∅})$

$→\frac{(sin∅+cos∅+1)(sin∅+cos∅-1)}{sin∅.cos∅}$

$→\frac{(sin∅+cos∅)²-(1²)}{sin∅.cos∅}$

$→\frac{sin²∅+cos²∅+2sin∅.cos∅-1}{sin∅.cos∅}$

$→\frac{1+2sin∅.cos∅-1}{sin∅.cos∅}$

$→\frac{2sin∅.cos∅}{sin∅.cos∅}$

here, both sin∅.cos∅ gets cancelled

$⇒\huge\boxed{2}$

Hence proved.....

Answer 2

Step-by-step explanation:

tanA−sinA

=

secA+1

secA−1

Taking LHS

tanA+sinA

tanA−sinA

=

cosA

sinA

+sinA

cosA

sinA

−sinA

=

sinAsecA+sinA

sinAsecA−sinA

=

secA+1

secA−1

LHS=RHS

Hence proved.

(i)

(1+tanA+secA)(1+cotA−cosecA)

(1+

cosA

sinA

+

cosA

1

)(1+

sinA

cosA

−

sinA

1

)

=(

cosA

cosA+sinA+1

)(

sinA

cosA+sinA−1

)

=

sinAcosA

(cosA+sinA)

2

−1

=

sinAcosA

sin

2

A+cos

2

A+2sinAcosA−1

=

sinAcosA

1+2sinAcosA−1

=2