Question

(sec²θ+cot²θ-Tan²θ) is equal to
​

Answer 1

Answer:

1

Step-by-step explanation:

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Answer 2

Answer:

Answer:

1

Step-by-step explanation:

Given : {\sf{\ \ {\dfrac{tan^2 \theta - sec^2 \theta}{cot^2 \theta - cosec^2 \theta}}}}

cot

2

θ−cosec

2

θ

tan

2

θ−sec

2

θ

____________________________

• {\boxed{\sf{\gray{1 + tan^2 \theta = sec^2 \theta}}}}

1+tan

2

θ=sec

2

θ

{\sf{\gray{From \ this, \ we \ get \ [ tan^2 \theta - sec^2 \theta = - 1]}}}From this, we get [tan

2

θ−sec

2

θ=−1]

____________________________

• {\boxed{\sf{\gray{1 + cot^2 \theta = cosec^2 \theta}}}}

1+cot

2

θ=cosec

2

θ

{\sf{\gray{From \ this, \ we \ get \ [ cot^2 \theta - cosec^2 \theta = - 1]}}}From this, we get [cot

2

θ−cosec

2

θ=−1]

____________________________

Putting known values, we get

____________________________

\Rightarrow{\sf{ {\dfrac{- 1}{- 1}} }}⇒

−1

−1

____________________________

\Rightarrow{\boxed{\sf{\red{1}}}}⇒

1

____________________________

Other identities which are frequently used :

• sin²θ + cos²θ = 1

• sin θ = P/H = 1/cosec θ

• cos θ = B/H = 1/sec θ

• tan θ = P/B = 1/cot θ

Step-by-step explanation:

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