Prove the following. \[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]

\[\frac{\tan\theta}{\sec\theta + 1}\] \[ = \frac{\tan\theta}{\sec\theta + 1} \times \frac{\sec\theta - 1}{\sec\theta - 1}\] \[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\sec^2 \theta - 1}\] \[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \] ...(1 + tan2θ = sec2θ) \[ = \frac{\sec\theta - 1}{\tan\theta}\]

Prove the following. tan⁡θsec⁡θ+1=sec⁡θ−1tan⁡θ - Geometry Mathematics 2

प्रश्न

Prove the following.

$\frac{\tan θ}{\sec θ + 1} = \frac{\sec θ - 1}{\tan θ}$

योग

उत्तर

$\frac{\tan θ}{\sec θ + 1}$

$= \frac{\tan θ}{\sec θ + 1} \times \frac{\sec θ - 1}{\sec θ - 1}$

$= \frac{\tan θ (\sec θ - 1)}{\sec^{2} θ - 1}$

$= \frac{\tan θ (\sec θ - 1)}{\tan^{2} θ}$ ...(1 + tan²θ = sec²θ)

$= \frac{\sec θ - 1}{\tan θ}$

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Application of Trigonometry

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 5.08 Q 5.07 Q 5.09

अध्याय 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board

अध्याय 6 Trigonometry
Problem Set 6 | Q 5.08 | पृष्ठ १३८

संबंधित प्रश्न

If $\sin θ = \frac{7}{25}$ , find the values of cosθ and tanθ.

If $\tan θ = \frac{3}{4}$ , find the values of secθ and cosθ

Prove that:

\sqrt{\frac{1 - \sin θ}{1 + \sin θ}} = \sec θ - \tan θ

Prove that:

\cot θ + \tan θ = c o s e c θ \sec θ

Prove that: $\frac{1}{sec θ − tan θ} = sec θ + tan θ$

Prove that:
If $\tan θ + \frac{1}{\tan θ} = 2$ , then show that $\tan^{2} θ + \frac{1}{\tan^{2} θ} = 2$

Prove that:

\sec^{4} A (1 - \sin^{4} A) - 2 \tan^{2} A = 1

Choose the correct alternative answer for the following question.
sin $θ$ cosec $θ$ = ?

Choose the correct alternative answer for the following question.

1 + tan² $θ$ = ?

Prove the following.

secθ (1 – sinθ) (secθ + tanθ) = 1

Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ

Prove the following.
cot²θ – tan²θ = cosec²θ – sec²θ

Prove the following.

\frac{1}{1 - \sin θ} + \frac{1}{1 + \sin θ} = 2 \sec^{2} θ

Prove the following.

\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ

Choose the correct alternative:
sinθ × cosecθ =?

ΔAMT∼ΔAHE, construct Δ AMT such that MA = 6.3 cm, ∠MAT=120°, AT = 4.9 cm and $\frac{MA}{HA} = \frac{7}{5}$ , then construct ΔAHE.

Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= $(\frac{1}{□} - \cos θ) (\frac{□}{□} + \frac{□}{□})$ ...... $[∵ \sec θ = \frac{1}{□}, \cot θ = \frac{□}{□} and \tan θ = \frac{□}{□}]$

= $(\frac{1 - □}{□}) (\frac{□ + □}{□ □})$

= $\frac{□}{□} \times \frac{1}{□ □}$ ...... $[\begin{matrix} ∵ □ + □ = 1 \\ ∴ □ = 1 - □ \end{matrix}]$

= $\frac{□}{□ □}$

= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Prove the following. tan⁡θsec⁡θ+1=sec⁡θ−1tan⁡θ - Geometry Mathematics 2

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Prove the following. tan⁡θsec⁡θ+1=sec⁡θ−1tan⁡θ - Geometry Mathematics 2

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