Question

If cot2θ (1 – 3 sec θ + 2 sec2θ) = 1, then 'θ' can be :

Answer 1

Step-by-step explanation:

The second and third identities can be obtained by manipulating the first. The identity 1+cot2θ=csc2θ1+cot2θ=csc2θ is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: 1+cot2θ=csc2θ1+cot2θ=csc2θ

Answer 2

Answer:

$Given, cot </p><p>2</p><p> θ( </p><p>1+sinθ</p><p>secθ−1</p><p> </p><p> )+sec </p><p>2</p><p> θ( </p><p>1+secθ</p><p>sinθ−1</p><p> </p><p> )</p><p></p><p>= </p><p>sinθ </p><p>2</p><p> </p><p>cosθ </p><p>2</p><p> </p><p> </p><p> × </p><p>(1+sinθ)</p><p>( </p><p>cosθ</p><p>1−cosθ</p><p> </p><p> )</p><p> </p><p> + </p><p>cos </p><p>2</p><p> θ</p><p>1</p><p> </p><p> </p><p>⎝</p><p>⎜</p><p>⎜</p><p>⎛</p><p> </p><p> </p><p>cosθ</p><p>cosθ+1</p><p> </p><p> </p><p>sinθ−1</p><p> </p><p> </p><p>⎠</p><p>⎟</p><p>⎟</p><p>⎞</p><p> </p><p> </p><p></p><p>= </p><p>(1−cos </p><p>2</p><p> θ)</p><p>cosθ</p><p> </p><p> × </p><p>(1+sinθ)</p><p>(1−cosθ)</p><p> </p><p> + </p><p>cosθ</p><p>1</p><p> </p><p> ( </p><p>cosθ+1</p><p>sinθ−1</p><p> </p><p> )</p><p></p><p>= </p><p>(1+cosθ)</p><p>cosθ</p><p> </p><p> × </p><p>(1+sinθ)</p><p>1</p><p> </p><p> + </p><p>cosθ(1+cosθ)</p><p>sinθ−1</p><p> </p><p> </p><p></p><p>= </p><p>(1+cosθ)(1+sinθ)cosθ</p><p>cosθ </p><p>2</p><p> +(1+sinθ)(sinθ−1)</p><p> </p><p> </p><p></p><p>= </p><p>(1+cosθ)(1+sinθ)cosθ</p><p>cosθ </p><p>2</p><p> +sinθ </p><p>2</p><p> −1</p><p> </p><p> =0.$

Step-by-step explanation:

Given, cot

2

θ(

1+sinθ

secθ−1

)+sec

2

θ(

1+secθ

sinθ−1

)

=

sinθ

2

cosθ

2

×

(1+sinθ)

(

cosθ

1−cosθ

)

+

cos

2

θ

1

⎝

⎜

⎜

⎛

cosθ

cosθ+1

sinθ−1

⎠

⎟

⎟

⎞

=

(1−cos

2

θ)

cosθ

×

(1+sinθ)

(1−cosθ)

+

cosθ

1

(

cosθ+1

sinθ−1

)

=

(1+cosθ)

cosθ

×

(1+sinθ)

1

+

cosθ(1+cosθ)

sinθ−1

=

(1+cosθ)(1+sinθ)cosθ

cosθ

2

+(1+sinθ)(sinθ−1)

=

(1+cosθ)(1+sinθ)cosθ

cosθ

2

+sinθ

2

−1

=0.