Question

(1+tan thita + sec thita ) ( 1+cot thita - cosec thita)=?​

Answer 1

$\bf\huge\blue{\underline{\underline{ Question : }}}$

(1+tan θ + sec θ ) ( 1+cot θ - cosec θ)=?

$\bf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

(1 + tan θ + sec θ)(1 + cot θ - cosec θ) = ?

★ Let,

LHS = (1 + tan θ + sec θ)(1 + cot θ - cosec θ)

$\sf\:\implies (1 + \tan\theta + \sec\theta)(1 + \cot\theta - \cosec\theta)$

• tan θ = sin θ/cos θ
• sec θ = 1/cos θ
• cot θ = cos θ/sin θ
• cosec θ = 1/sin θ

$\sf\:\implies (1 + \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta})(1 + \frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta})$

$\sf\:\implies (\frac{(\cos\theta + \sin\theta) + 1}{\cos\theta})(\frac{(\sin\theta+\cos\theta)-1}{\sin\theta})$

• [ (a + b)(a - b) = a² - b² ]

$\sf\:\implies (\frac{(\cos\theta + \sin\theta)^{2} - (1)^{2}}{\cos\theta\sin\theta})$

• (a + b)² = a² + b² + 2ab

$\sf\:\implies (\frac{\cos^{2}\theta + \sin^{2}\theta + 2\sin\theta\cos\theta - 1}{\cos\theta\sin\theta}$

• sin² θ + cos² θ = 1

$\sf\:\implies (\frac{1 + 2\sin\theta\cos\theta - 1}{\cos\theta\sin\theta})$

$\sf\:\implies \frac{2 \sin\theta\cos\theta}{\cos\theta\sin\theta}$

$\sf\:\implies 2$

RHS = " 2 "

$\underline{\boxed{\bf{\purple{ \therefore (1 + \tan\theta+ \sec\theta)(1 + \cot\theta-\cosec\theta) = 2}}}}\:\orange{\bigstar}$

★ More Information :

$\boxed{\begin{minipage}{7 cm} Fundamental Trigonometric Identities : \\ \\ \sin^{2}\theta + cos^{2}\theta = 1 \\ \\ 1 + tan^{2}\theta = sec^{2}\theta \\ \\1 + cot^{2}\theta=\text{cosec}^2\, \theta \end{minipage}}$