Solveeit Logo
Login

Question

Mathematics Question on Trigonometric Identities

Choose the correct option. Justify your choice. (1+tan θ+sec θ)(1+cot θ - cosec θ)=(1+\text{tan θ+sec θ}) (1+\text{cot θ - cosec θ}) =

A

0

B

1

C

2

D

-1

Answer

2

Explanation

Solution

(1+tan θ+sec θ)(1+cot θ - cosec θ)(1+\text{tan θ+sec θ}) (1+\text{cot θ - cosec θ})
We know,
tan (x) = sin (x)cos (x)\frac{\text{sin (x)}}{\text{cos (x)}}

cot (x) =cos (x)sin (x)=1tan (x) \frac{\text{cos (x)}}{\text{sin (x)}} = \frac{1}{\text{tan (x)}}

sec (x) = 1cos (x)\frac{1}{\text{cos (x)}}

cosec (x) = 1sin (x)\frac{1}{\text{sin (x)}}

By substituting
=[(1+sin θcos θ+1cos θ)(1+cos θsin θ −1sin θ)]= \left[\left(1 + \frac{\text{sin θ}}{\text{cos θ}} + \frac{1}{\text{cos θ}}\right) \left(1 + \frac{\text{cos θ}}{\text{sin θ }}- \frac{1}{\text{sin θ}}\right)\right]

=\left [\frac{(\text{cos θ + sin θ }+ 1)}{\text{cos θ}}\right]$$\left [\frac{(\text{sin θ + cos θ }- 1)}{\text{sin θ}}\right] (By taking LCM and multiplying)

=[(sin θ + cos θ)2\-(−1)2]sin θ cos θ = \frac{\left[(\text{sin θ + cos θ})^2 \- (- 1) ^2\right] }{ \text{sin θ cos θ }} [Using a2 - b2 = (a + b) (a - b)]

=sin2θ+cos2θ+2sin θ cos θ −1sin θ cos θ= \frac{\text{sin}^2 \text{θ} + \text{cos}^2 \text{θ} + 2 \text{sin θ cos θ }- 1 }{\text{sin θ cos θ}}

=(1+2 sin θ cos θ −1) sin θ cos θ= \frac{(1 + 2\ \text{sin θ cos θ }- 1) }{\text{ sin θ cos θ}} (Using identity sin2 θ + cos2 θ = 1)

=2 sin θ cos θ  sin θ cos θ=\frac{ 2\ \text{sin θ cos θ }}{\text{ sin θ cos θ}}
= 2
Hence, alternative (C) is correct.