# 2 3 C O S E C 2 58 ∘ − 2 3 Cot 58 ∘ Tan 32 ∘ − 5 3 Tan 13 ∘ Tan 37 ∘ Tan 45 ∘ Tan 53 ∘ Tan 77 ∘ = − 1 - Mathematics

Sum
$\frac{2}{3} {cosec}^2 58^\circ- \frac{2}{3}\cot58^\circ \tan32^\circ - \frac{5}{3}\tan13^\circ \tan37^\circ\tan45^\circ\tan53^\circ\tan77^\circ = - 1$

#### Solution

$\begin{array}{l}\frac{2}{3}\cos {ec}^2 {58}^0 - \frac{2}{3}\cot {58}^0 \tan {32}^0 - \frac{5}{3}\tan {13}^0 \tan {37}^0 \tan {45}^0 \tan {53}^0 \tan {77}^0 \\ \end{array}$

$\begin{array}{l}=\frac{2}{3}(\cos {ec}^2 {58}^0 - \cot {58}^0 \tan {32}^0 ) - \frac{5}{3}\tan {13}^0 \tan( {90}^0 - {13}^0 )\tan {37}^0 \tan( {90}^0 - {37}^0 )(\tan {45}^0 ) \\ \end{array}$

$\begin{array}{l}=\frac{2}{3}{\cos {ec}^2 {58}^0 - \cot {58}^0 \tan( {90}^0 - {58}^0 )}- \frac{5}{3}\tan {13}^0 \cot {13}^0 \tan {37}^0 \cot {37}^0 (1) \\ \end{array}$

$\begin{array}{l}=\frac{2}{3}(\cos {ec}^2 {58}^0 - \cot {58}^0 \cot {58}^0 ) - \frac{5}{3}\tan {13}^0 \frac{1}{\tan {13}^0}\tan {37}^0 \frac{1}{\tan {37}^0} \\ \end{array}$

$\begin{array}{l}=\frac{2}{3}(\cos {ec}^2 {58}^0 - \cot^2 {58}^0 ) - \frac{5}{3} \\ \end{array}$
$\begin{array}{l}=\frac{2}{3} - \frac{5}{3} \\ \end{array}$

= -1

Hence Proved

Concept: Trigonometry
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#### APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 15 | Page 314

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