Respuesta :

#33

LHS

[tex]\\ \rm\Rrightarrow \dfrac{sin\theta}{csc\theta}+\dfrac{cos\theta}{sec\theta}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sin\theta}{\dfrac{1}{sin\theta}}+\dfrac{cos\theta}{\dfrac{1}{cos\theta}}[/tex]

[tex]\\ \rm\Rrightarrow sin^2\theta+cos^2\theta[/tex]

[tex]\\ \rm\Rrightarrow 1[/tex]

  • Proved

#34

  • Ø is taken as A for easy typing

LHS

[tex]\\ \rm\Rrightarrow tanAcsc^2A-tanA[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sinA}{cosA}\times \dfrac{1}{sin^2A}-\dfrac{sinA}{cosA}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sinA}{sin^2AcosA}-\dfrac{sinA}{cosA}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{sinAcosA}-\dfrac{sinA}{cosA}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1-sin^2A}{sinAcosA}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{cos^2A}{sinAcosA}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{cosA}{sinA}[/tex]

[tex]\\ \rm\Rrightarrow cotA[/tex]

Proved

semsee45

Answer:

Trigonometric Identities used:

[tex]\csc \theta=\dfrac{1}{\sin \theta}[/tex]

[tex]\sec \theta=\dfrac{1}{\cos \theta}[/tex]

[tex]\cot \theta=\dfrac{1}{\tan \theta}[/tex]

[tex]\sin^2 \theta + \cos^2 \theta=1[/tex]

[tex]\csc^2 \theta = 1 + \cot^2 \theta[/tex]

Question 33

[tex]\large \begin{aligned}\dfrac{\sin \theta}{\csc \theta}+\dfrac{\cos \theta}{\sec \theta}& = \sin \theta \cdot \dfrac{1}{\csc \theta}+\cos \theta \cdot \dfrac{1}{\sec \theta}\\\\& = \sin \theta \cdot \dfrac{1}{\frac{1}{\sin \theta}}+\cos \theta \cdot \dfrac{1}{\frac{1}{\cos \theta}}\\\\& = \sin \theta \cdot \sin \theta+\cos \theta \cdot \cos \theta\\\\& = \sin^2 \theta + \cos^2 \theta\\\\& = 1\end{aligned}[/tex]

Question 34

[tex]\large \begin{aligned}\tan \theta \csc^2 \theta - \tan \theta & = \tan \theta ( \csc^2 \theta - 1)\\\\& = \tan \theta (1 + \cot^2 \theta -1)\\\\& = \tan \theta \cot^2 \theta \\\\& = \dfrac{1}{\cot \theta} \cdot \cot^2 \theta \\\\& = \dfrac{\cot^2 \theta}{\cot \theta}\\\\& = \cot \theta\end{aligned}[/tex]

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