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प्रश्न
Prove the following trigonometric identities:
`sqrt(("cosec" θ - 1)/("cosec" θ + 1)) + sqrt(("cosec" θ + 1)/("cosec" θ - 1)) = 2 sec θ`
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उत्तर
Given: = `sqrt(("cosec" θ - 1)/("cosec" θ + 1)) + sqrt(("cosec" θ + 1)/("cosec" θ - 1))`
To Prove: `sqrt(("cosec" θ − 1)/("cosec" θ + 1)) + sqrt(("cosec" θ + 1)/("cosec" θ − 1)) = 2 sec θ`
Proof [Step-wise]:
1. Let a = cosec θ.
Consider the left-hand side (LHS):
LHS = `sqrt((a - 1)/(a + 1)) + sqrt((a + 1)/(a - 1))`
2. Write each term with common radical denominator:
= `sqrt(a - 1)/sqrt(a + 1) + sqrt(a + 1)/sqrt(a - 1)`
3. Multiply numerator and denominator of each term to get a common denominator `sqrt(a^2 - 1)`:
= `(a - 1)/sqrt(a^2 - 1) + (a + 1)/sqrt(a^2 - 1)`
= `((a - 1) + (a + 1))/sqrt(a^2 - 1)`
= `(2a)/sqrt(a^2 - 1)`
4. Substitute back a = cosec θ:
LHS = `(2 "cosec" θ)/sqrt("cosec"^2θ - 1)`
5. Simplify the radical:
`sqrt("cosec"^2θ - 1)`
= `sqrt((1/sin^2θ) - 1)`
= `sqrt(cos^2θ/sin^2θ)`
= |cot θ|
Hence LHS = `(2 "cosec" θ)/|cot θ|`
= `2/|cos θ|`
= 2 |sec θ|
6. Therefore, for angles where the expression is defined, LHS = 2 |sec θ|.
If we restrict to quadrants where cos θ > 0 (so |sec θ| = sec θ), then LHS = 2 sec θ.
`sqrt(("cosec" θ - 1)/("cosec" θ + 1)) + sqrt(("cosec" θ + 1)/("cosec" θ - 1)) = 2 |sec θ|` in general for θ where the expression is defined. Under the usual sign assumption cos θ > 0 (so sec θ > 0), this reduces to the stated identity 2 sec θ.
